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Mathematics LibreTexts

7-3. Double-Angle, Half-Angle, and Reduction Formulas

Double-Angle, Half-Angle, and Reduction Formulas
In this section, you will:
  • Use double-angle formulas to find exact values.
  • Use double-angle formulas to verify identities.
  • Use reduction formulas to simplify an expression.
  • Use half-angle formulas to find exact values.
<figure id="Figure_07_03_001" style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: auto; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 1; word-spacing: 0px; -webkit-text-stroke-width: 0px;"> <figcaption>Bicycle ramps for advanced riders have a steeper incline than those designed for novices.</figcaption> Picture of two bicycle ramps, one with a steep slope and one with a gentle slope.</figure>

Bicycle ramps made for competition (see [link]) must vary in height depending on the skill level of the competitors. For advanced competitors, the angle formed by the ramp and the ground should be<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>such that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 5 3 . The angle is divided in half for novices. What is the steepness of the ramp for novices? In this section, we will investigate three additional categories of identities that we can use to answer questions such as this one.

Using Double-Angle Formulas to Find Exact Values

In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>α</mi><mo>=</mo><mi>β</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics></math>Deriving the double-angle formula for sine begins with the sum formula,

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> α+β )=sin α cos β+cos α sin β

If we let<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>α</mi><mo>=</mo><mi>β</mi><mo>=</mo><mi>θ</mi><mo>,</mo></mrow></annotation-xml></semantics></math>then we have

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> θ+θ )=sin θ cos θ+cos θ sin θ     sin( 2θ )=2sin θ cos θ

Deriving the double-angle for cosine gives us three options. First, starting from the sum formula,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> α+β )=cos α cos β−sin α sin β, and letting<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>α</mi><mo>=</mo><mi>β</mi><mo>=</mo><mi>θ</mi><mo>,</mo></mrow></annotation-xml></semantics></math> we have

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>cos</mi><mo stretchy="false">(</mo><mi>θ</mi><mo>+</mo><mi>θ</mi><mo stretchy="false">)</mo><mo>=</mo><mi>cos</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>−</mo><mi>sin</mi><mtext> </mtext><mi>θ</mi><mi>sin</mi><mtext> </mtext><mi>θ</mi></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math>     cos(2θ)= cos 2 θ− sin 2 θ

Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> cos 2 θ− sin 2 θ             =(1− sin 2 θ)− sin 2 θ             =1−2 sin 2 θ

The second interpretation is:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> cos 2 θ− sin 2 θ             = cos 2 θ−(1− cos 2 θ)             =2  cos 2 θ−1

Similarly, to derive the double-angle formula for tangent, replacing<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>α</mi><mo>=</mo><mi>β</mi><mo>=</mo><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>in the sum formula gives

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable><mtr><mtd><mrow><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> α+β )= tan α+tan β 1−tan α tan β tan( θ+θ )= tan θ+tan θ 1−tan θ tan θ tan( 2θ )= 2tan θ 1− tan 2 θ
Double-Angle Formulas

The double-angle formulas are summarized as follows:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2θ )=2 sin θ cos θ
 
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> cos 2 θ− sin 2 θ             =1−2  sin 2 θ             =2  cos 2 θ−1
 
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2θ )= 2 tan θ 1− tan 2 θ

Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.

  1. Draw a triangle to reflect the given information.
  2. Determine the correct double-angle formula.
  3. Substitute values into the formula based on the triangle.
  4. Simplify.
Using a Double-Angle Formula to Find the Exact Value Involving Tangent

Given that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> 3 4  and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is in quadrant II, find the following:

  1. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2θ )
  2. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2θ )
  3. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2θ )

If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> 3 4 ,such that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is in the second quadrant, the adjacent side is on the x-axis and is negative. Use the Pythagorean Theoremto find the length of the hypotenuse:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="right"><mtr columnalign="right"><mtd columnalign="right"><mrow><msup><mrow><mo stretchy="false">(−4</mo><mo stretchy="false">)</mo></mrow></msup></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> 2 + (3) 2 = c 2 16+9= c 2 25= c 2 c=5  

Now we can draw a triangle similar to the one shown in [link].

<figure class="small" id="Figure_07_03_02">Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-4,0), and (-4,3). The angle at the origin is theta. The angle formed by the side (-4,3) to (-4,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 5.</figure>
  1. Let’s begin by writing the double-angle formula for sine.
    <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>sin</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo><mo>=</mo><mn>2</mn><mtext> </mtext><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>θ</mi></mrow></annotation-xml></semantics></math>

    We see that we to need to find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics></math>Based on [link], we see that the hypotenuse equals 5, so<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 3 5 , and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> 4 5 . Substitute these values into the equation, and simplify.

    Thus,

    <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>sin</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo><mo>=</mo><mn>2</mn><mrow><mo>(</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> 3 5 )( − 4 5 )             =− 24 25
  2. Write the double-angle formula for cosine.
    <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2θ )= cos 2 θ− sin 2 θ

    Again, substitute the values of the sine and cosine into the equation, and simplify.

    <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> ( − 4 5 ) 2 − ( 3 5 ) 2             = 16 25 − 9 25             = 7 25
  3. Write the double-angle formula for tangent.
    <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>tan</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 2 tan θ 1− tan 2 θ

    In this formula, we need the tangent, which we were given as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> 3 4 . Substitute this value into the equation, and simplify.

    <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>tan</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> 2( − 3 4 ) 1− ( − 3 4 ) 2            = − 3 2 1− 9 16            =− 3 2 ( 16 7 )            =− 24 7

Given<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mtext> </mtext><mi>α</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 5 8 , with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>in quadrant I, find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2α ).

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2α )= 7 32

Using the Double-Angle Formula for Cosine without Exact Values

Use the double-angle formula for cosine to write<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 6x ) in terms of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 3x ).

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>cos</mi><mo stretchy="false">(</mo><mn>6</mn><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mi>cos</mi><mo stretchy="false">(</mo><mn>3</mn><mi>x</mi><mo>+</mo><mn>3</mn><mi>x</mi><mo stretchy="false">)</mo></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math>             =cos 3x cos 3x−sin 3x sin 3x             = cos 2 3x− sin 2 3x
Analysis

This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.

Using Double-Angle Formulas to Verify Identities

Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.

Using the Double-Angle Formulas to Establish an Identity

Establish the following identity using double-angle formulas:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>1</mn><mo>+</mo><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2θ )= ( sin θ+cos θ ) 2

We will work on the right side of the equal sign and rewrite the expression until it matches the left side.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msup><mrow><mo stretchy="false">(</mo><mi>sin</mi><mtext> </mtext><mi>θ</mi><mo>+</mo><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo stretchy="false">)</mo></mrow></msup></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> 2 = sin 2 θ+2 sin θ cos θ+ cos 2 θ                        =( sin 2 θ+ cos 2 θ)+2 sin θ cos θ                        =1+2 sin θ cos θ                       =1+sin(2θ)
Analysis

This process is not complicated, as long as we recall the perfect square formula from algebra:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></annotation-xml></semantics></math> a±b ) 2 = a 2 ±2ab+ b 2

where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>=</mo><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>=</mo><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics></math>Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent.

Establish the identity:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics></math> cos 4 θ− sin 4 θ=cos( 2θ ).

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>cos</mi></mrow></msup></mrow></annotation-xml></semantics></math> 4 θ− sin 4 θ=( cos 2 θ+ sin 2 θ )( cos 2 θ− sin 2 θ )=cos( 2θ )

Verifying a Double-Angle Identity for Tangent

Verify the identity:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2θ )= 2 cot θ−tan θ

In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> 2θ )= 2 tan θ 1− tan 2 θ Double-angle formula            = 2 tan θ( 1 tan θ ) ( 1− tan 2 θ )( 1 tan θ )Multiply by a term that results in desired numerator.            = 2 1 tan θ − tan 2 θ tan θ            = 2 cot θ−tan θUse reciprocal identity for  1 tan θ .
Analysis

Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>2</mn><mi>tan</mi><mtext> </mtext><mi>θ</mi></mrow></mfrac></mrow></annotation-xml></semantics></math> 1− tan 2 θ = 2 cot θ−tan θ

Let’s work on the right side.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mfrac><mn>2</mn></mfrac></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> cot θ−tan θ = 2 1 tan θ −tan θ ( tan θ tan θ )                    = 2 tan θ 1 tan θ ( tan θ )−tan θ(tan θ)                    = 2 tan θ 1− tan 2 θ

When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier.

Verify the identity:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><msup/></mrow></annotation-xml></semantics></math> cos 3 θ−cos θ  sin 2 θ.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2θ )cos θ=( cos 2 θ− sin 2 θ )cos θ= cos 3 θ−cos θ sin 2 θ

Use Reduction Formulas to Simplify an Expression

The double-angle formulas can be used to derive the reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.

We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2θ )=1−2  sin 2 θ. Solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics></math> sin 2 θ:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn><mo>−</mo><mn>2</mn><mtext> </mtext><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> sin 2 θ  2  sin 2 θ=1−cos(2θ)     sin 2 θ= 1−cos(2θ) 2

Next, we use the formula<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2θ )=2  cos 2 θ−1. Solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics></math> cos 2 θ:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>       </mtext><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo><mo>=</mo><mn>2</mn><mtext> </mtext><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> cos 2 θ−1  1+cos(2θ)=2  cos 2 θ 1+cos(2θ) 2 = cos 2 θ

The last reduction formula is derived by writing tangent in terms of sine and cosine:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msup><mrow><mi>tan</mi></mrow></msup></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> 2 θ= sin 2 θ cos 2 θ          = 1−cos(2θ) 2 1+cos(2θ) 2 Substitute the reduction formulas.          =( 1−cos(2θ) 2 )( 21+cos(2θ) )          = 1−cos(2θ) 1+cos(2θ)
Reduction Formulas

The reduction formulas are summarized as follows:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>sin</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 θ= 1−cos( 2θ ) 2
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>cos</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 θ= 1+cos( 2θ ) 2
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>tan</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 θ= 1−cos( 2θ ) 1+cos( 2θ )
Writing an Equivalent Expression Not Containing Powers Greater Than 1

Write an equivalent expression for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics></math> cos 4 x that does not involve any powers of sine or cosine greater than 1.

We will apply the reduction formula for cosine twice.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msup><mrow><mi>cos</mi></mrow></msup></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> 4 x= ( cos 2 x) 2           = ( 1+cos(2x) 2 ) 2 Substitute reduction formula for cos 2 x.           = 1 4 ( 1+2cos(2x)+ cos 2 (2x))           = 1 4 + 1 2  cos(2x)+ 1 4 ( 1+cos2(2x) 2 )  Substitute reduction formula for cos 2 x.           = 1 4 + 1 2  cos(2x)+ 1 8 +1 8  cos(4x)           = 3 8 + 1 2  cos(2x)+ 1 8  cos(4x)
Analysis

The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra.

Using the Power-Reducing Formulas to Prove an Identity

Use the power-reducing formulas to prove

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>sin</mi></mrow></msup></mrow></annotation-xml></semantics></math> 3 ( 2x )=[ 1 2  sin( 2x ) ] [ 1−cos( 4x ) ]

We will work on simplifying the left side of the equation:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msup><mrow><mi>sin</mi></mrow></msup></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> 3 (2x)=[sin(2x)][ sin 2 (2x)]              =sin(2x)[ 1−cos(4x) 2 ] Substitute the power-reduction formula.              =sin(2x)( 12 )[ 1−cos(4x) ]              = 1 2 [sin(2x)][1−cos(4x)]
Analysis

Note that in this example, we substituted

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>1</mn><mo>−</mo><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></mfrac></mrow></annotation-xml></semantics></math> 4x ) 2

for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics></math> sin 2 ( 2x ). The formula states

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>sin</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 θ= 1−cos( 2θ ) 2

We let<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mo>=</mo><mn>2</mn><mi>x</mi><mo>,</mo></mrow></annotation-xml></semantics></math>so<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mi>θ</mi><mo>=</mo><mn>4</mn><mi>x</mi><mo>.</mo></mrow></annotation-xml></semantics></math>

Use the power-reducing formulas to prove that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>10</mn><mtext> </mtext><msup/></mrow></annotation-xml></semantics></math> cos 4 x= 15 4 +5 cos( 2x )+ 5 4  cos( 4x ).

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>10</mn><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> cos 4 x=10 cos 4 x=10 ( cos 2 x) 2             =10 [ 1+cos(2x) 2 ] 2 Substitute reduction formula for cos 2 x.             = 10 4[1+2cos(2x)+ cos 2 (2x)]             = 10 4 + 10 2 cos(2x)+ 10 4 ( 1+cos2(2x) 2 ) Substitute reduction formula for cos 2 x.            = 10 4 + 10 2 cos(2x)+ 10 8 + 10 8 cos(4x)             = 30 8 +5cos(2x)+ 10 8 cos(4x)             = 15 4 +5cos(2x)+ 5 4cos(4x)

Using Half-Angle Formulas to Find Exact Values

The next set of identities is the set of half-angle formulas, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> α 2 , the half-angle formula for sine is found by simplifying the equation and solving for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> α 2 ). Note that the half-angle formulas are preceded by a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo>±</mo><mtext> </mtext></mrow></annotation-xml></semantics></math>sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> α 2  terminates.

The half-angle formula for sine is derived as follows:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>   </mtext><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> sin 2 θ= 1−cos(2θ) 2 sin 2 ( α 2 )= 1−( cos2⋅ α 2 ) 2              = 1−cos α 2    sin( α 2 )=± 1−cos α 2

To derive the half-angle formula for cosine, we have

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>   </mtext><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> cos 2 θ= 1+cos(2θ) 2 cos 2 ( α 2 )= 1+cos( 2⋅ α 2 ) 2              = 1+cos α 2   cos( α 2 )=± 1+cos α 2

For the tangent identity, we have

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>   </mtext><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> tan 2 θ= 1−cos(2θ) 1+cos(2θ) tan 2 ( α 2 )= 1−cos( 2⋅ α 2 ) 1+cos( 2⋅ α 2 )             = 1−cos α 1+cos α   tan( α 2 )=± 1−cos α 1+cos α
Half-Angle Formulas

The half-angle formulas are as follows:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> α 2 )=± 1−cos α 2
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> α 2 )=± 1+cos α 2
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> α 2 )=± 1−cos α 1+cos α                        = sin α 1+cos α                        = 1−cos α sin α
Using a Half-Angle Formula to Find the Exact Value of a Sine Function

Find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 15 ∘ ) using a half-angle formula.

Since<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics></math> 15 ∘ = 30 ∘ 2 , we use the half-angle formula for sine:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>sin</mi><mtext> </mtext><mfrac/></mrow></mtd></mtr></mtable></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> 30 ∘ 2 = 1−cos 30 ∘ 2            = 1− 3 2 2            = 2− 3 2 2            = 2− 3 4            = 2− 3 2
Analysis

Notice that we used only the positive root because<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mo stretchy="false">(</mo><msup/></mrow></annotation-xml></semantics></math> 15 o ) is positive.

Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.

  1. Draw a triangle to represent the given information.
  2. Determine the correct half-angle formula.
  3. Substitute values into the formula based on the triangle.
  4. Simplify.
Finding Exact Values Using Half-Angle Identities

Given that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mtext> </mtext><mi>α</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 8 15 and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>α</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>lies in quadrant III, find the exact value of the following:

  1. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> α 2 )
  2. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> α 2 )
  3. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> α 2 )

Using the given information, we can draw the triangle shown in [link]. Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mtext> </mtext><mi>α</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> 8 17  and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>α</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> 15 17 .

<figure class="medium" id="Figure_07_03_003">Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-15,0), and (-15,-8). The angle at the origin is alpha. The angle formed by the side (-15,-8) to (-15,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 17.</figure>
  1. Before we start, we must remember that, if<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>α</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is in quadrant III, then<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>180°</mn><mo><</mo><mi>α</mi><mo><</mo><mn>270°</mn><mo>,</mo></mrow></annotation-xml></semantics></math>so<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> 180° 2 < α 2 < 270° 2 . This means that the terminal side of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> α 2  is in quadrant II, since<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>90°</mn><mo><</mo><mfrac/></mrow></annotation-xml></semantics></math> α 2 <135°.

    To find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> α 2 , we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in [link] and simplify.

    <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mtable columnalign="left"><mtr><mtd><mrow/></mtd></mtr><mtr><mtd><mi>sin</mi><mtext> </mtext><mfrac/></mtd></mtr></mtable></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> α 2 =± 1−cos α 2         =± 1−( − 15 17 ) 2         =± 32 17 2         =± 32 17 ⋅ 1 2         =± 16 17         =± 4 17         = 417 17

    We choose the positive value of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> α 2  because the angle terminates in quadrant II and sine is positive in quadrant II.

  2. To find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> α 2 , we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in [link], and simplify.
    <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>cos</mi><mtext> </mtext><mfrac/></mrow></mtd></mtr></mtable></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> α 2 =± 1+cos α 2         =± 1+( − 15 17 ) 2         =± 2 17 2         =± 2 17 ⋅ 1 2         =± 1 17         =− 17 17

    We choose the negative value of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> α 2  because the angle is in quadrant II because cosine is negative in quadrant II.

  3. To find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> α 2 , we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle in [link] and simplify.
    <math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>tan</mi><mtext> </mtext><mfrac/></mrow></mtd></mtr></mtable></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> α 2 =± 1−cos α 1+cos α         =± 1−(− 15 17 ) 1+(− 15 17 )         =± 32 17 2 17         =± 32 2         =− 16         =−4

    We choose the negative value of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> α 2  because<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> α 2  lies in quadrant II, and tangent is negative in quadrant II.

Given that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mtext> </mtext><mi>α</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> 4 5  and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>α</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>lies in quadrant IV, find the exact value of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> α 2 ).

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> 2 5

Finding the Measurement of a Half Angle

Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 5 3  for higher-level competition, what is the measurement of the angle for novice competition?

Since the angle for novice competition measures half the steepness of the angle for the high level competition, and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 5 3  for high competition, we can find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>from the right triangle and the Pythagorean theorem so that we can use the half-angle identities. See [link].

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><msup><mn>3</mn></msup></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> 2 + 5 2 =34           c= 34
<figure class="small" id="Figure_07_03_004">Image of a right triangle with sides 3, 5, and rad34. Rad 34 is the hypotenuse, and 3 is the base. The angle formed by the hypotenuse and base is theta. The angle between the side of length 3 and side of length 5 is a right angle.</figure>

We see that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 3 34 = 3 34 34 . We can use the half-angle formula for tangent:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> θ 2 = 1−cos θ 1+cos θ . Since<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is in the first quadrant, so is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> θ 2 . Thus,

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>tan</mi><mtext> </mtext><mfrac/></mrow></mtd></mtr></mtable></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> θ 2 = 1− 3 34 34 1+ 3 34 34         = 34−3 34 34 34+3 34 34         = 34−3 34 34+3 34         ≈0.57

We can take the inverse tangent to find the angle:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics></math> tan −1 ( 0.57 )≈ 29.7 ∘ . So the angle of the ramp for novice competition is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo>≈</mo><msup/></mrow></annotation-xml></semantics></math> 29.7 ∘ .

Access these online resources for additional instruction and practice with double-angle, half-angle, and reduction formulas.

Key Equations

Double-angle formulas <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>sin</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo><mo>=</mo><mn>2</mn><mi>sin</mi><mtext> </mtext><mi>θ</mi><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>θ</mi></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> cos(2θ)= cos 2 θ− sin 2 θ            =1−2 sin 2 θ            =2 cos 2 θ−1 tan(2θ)= 2tan θ1− tan 2 θ
Reduction formulas <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><msup><mi>sin</mi></msup></mtd></mtr></mtable></annotation-xml></semantics></math> 2 θ= 1−cos( 2θ ) 2 cos 2 θ= 1+cos( 2θ ) 2 tan 2 θ= 1−cos( 2θ ) 1+cos( 2θ )
Half-angle formulas <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>sin</mi><mtext> </mtext><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> α 2 =± 1−cos α 2 cos  α 2 =± 1+cos α 2 tan  α 2 =± 1−cos α 1+cos α         = sin α 1+cos α         =1−cos α sin α

Key Concepts

  • Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent. See [link], [link], [link], and [link].
  • Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term. See [link] and [link].
  • Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not. See [link], [link], and [link].

Section Exercises

Verbal

Explain how to determine the reduction identities from the double-angle identity<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2x )= cos 2 x− sin 2 x.

Use the Pythagorean identities and isolate the squared term.

Explain how to determine the double-angle formula for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mo stretchy="false">(</mo><mn>2</mn><mi>x</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics></math>using the double-angle formulas for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>x</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics></math>and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mo stretchy="false">(</mo><mn>2</mn><mi>x</mi><mo stretchy="false">)</mo><mo>.</mo></mrow></annotation-xml></semantics></math>

We can determine the half-angle formula for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x 2 )= 1−cos x 1+cos x  by dividing the formula for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x 2 ) by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x 2 ). Explain how to determine two formulas for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x 2 )  that do not involve any square roots.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> 1−cos x sin x , sin x 1+cos x , multiplying the top and bottom by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msqrt/></mrow></annotation-xml></semantics></math> 1−cos x  and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msqrt/></mrow></annotation-xml></semantics></math> 1+cos x , respectively.

For the half-angle formula given in the previous exercise for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x 2 ),explain why dividing by 0 is not a concern. (Hint: examine the values of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>necessary for the denominator to be 0.)

Algebraic

For the following exercises, find the exact values of a)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2x ), b)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2x ), and c)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2x ) without solving for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>.</mo></mrow></annotation-xml></semantics></math>

If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mtext> </mtext><mi>x</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 1 8 , and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is in quadrant I.

a)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> 3 7 32  b)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> 31 32  c)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> 3 7 31

If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>x</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 2 3 , and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is in quadrant I.

If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>x</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> 1 2 , and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>  </mtext><mi>x</mi><mtext>  </mtext></mrow></annotation-xml></semantics></math>is in quadrant III.

a)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> 3 2  b)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> 1 2  c)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo>−</mo><msqrt/></mrow></annotation-xml></semantics></math> 3  

If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mtext> </mtext><mi>x</mi><mo>=</mo><mo>−</mo><mn>8</mn><mo>,</mo></mrow></annotation-xml></semantics></math> and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>  </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is in quadrant IV.

For the following exercises, find the values of the six trigonometric functions if the conditions provided hold.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 3 5  and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics></math> 90 ∘ ≤θ≤ 180 ∘

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mtext> </mtext><mi>θ</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> 2 5 5 ,sin θ= 5 5 ,tan θ=− 1 2 ,csc θ= 5 ,sec θ=− 5 2 ,cot θ=−2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 1 2  and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics></math> 180 ∘ ≤θ≤ 270 ∘

For the following exercises, simplify to one trigonometric expression.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>2</mn><mtext> </mtext><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> π 4 ) 2 cos( π 4 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>2</mn><mtext> </mtext><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> π 2 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>4</mn><mtext> </mtext><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> π 8 ) cos( π 8 )

For the following exercises, find the exact value using half-angle formulas.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> π 8 ) 

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msqrt><mrow><mn>2</mn><mo>−</mo><msqrt/></mrow></msqrt></mrow></mfrac></mrow></annotation-xml></semantics></math> 2 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> − 11π 12 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 11π 12 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msqrt><mrow><mn>2</mn><mo>−</mo><msqrt/></mrow></msqrt></mrow></mfrac></mrow></annotation-xml></semantics></math> 3 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 7π 8 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 5π 12 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>2</mn><mo>+</mo><msqrt/></mrow></annotation-xml></semantics></math> 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> − 3π 12 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> − 3π 8 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mn>1</mn><mo>−</mo><msqrt/></mrow></annotation-xml></semantics></math> 2

For the following exercises, find the exact values of a)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x 2 ), b)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x 2 ), and c)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x 2 )without solving for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>.</mo></mrow></annotation-xml></semantics></math>

If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mtext> </mtext><mi>x</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> 4 3 , and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>  </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is in quadrant IV.

If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mtext> </mtext><mi>x</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> 12 13 , and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is in quadrant III.

a)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> 3 13 13  b)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> 2 13 13  c)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> 3 2  

If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>csc</mi><mtext> </mtext><mi>x</mi><mo>=</mo><mn>7</mn><mo>,</mo></mrow></annotation-xml></semantics></math> and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext>  </mtext></mrow></annotation-xml></semantics></math>is in quadrant II.

If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sec</mi><mtext> </mtext><mi>x</mi><mo>=</mo><mo>−</mo><mn>4</mn><mo>,</mo></mrow></annotation-xml></semantics></math> and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is in quadrant II.

a)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> 10 4  b)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> 6 4  c)<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> 15 3  

For the following exercises, use [link] to find the requested half and double angles.

<figure class="small" id="Figure_07_03_201">Image of a right triangle. The base is length 12, and the height is length 5. The angle between the base and the height is 90 degrees, the angle between the base and the hypotenuse is theta, and the angle between the height and the hypotenuse is alpha degrees.</figure>

Find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2θ ),cos(2θ), and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mo stretchy="false">(</mo><mn>2</mn><mi>θ</mi><mo stretchy="false">).</mo></mrow></annotation-xml></semantics></math>

Find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mo stretchy="false">(</mo><mn>2</mn><mi>α</mi><mo stretchy="false">)</mo><mo>,</mo><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>α</mi><mo stretchy="false">)</mo><mo>,</mo></mrow></annotation-xml></semantics></math> and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mo stretchy="false">(</mo><mn>2</mn><mi>α</mi><mo stretchy="false">).</mo></mrow></annotation-xml></semantics></math>

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>120</mn></mrow></mfrac></mrow></annotation-xml></semantics></math> 169 ,– 119 169 ,– 120 119

Find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> θ 2 ),cos( θ 2 ), and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> θ 2 ).

Find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> α 2 ),cos( α 2 ), and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>tan</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> α 2 ).

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>2</mn><msqrt/></mrow></mfrac></mrow></annotation-xml></semantics></math> 13 13 , 3 13 13 , 2 3

For the following exercises, simplify each expression. Do not evaluate.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>cos</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 ( 28 ∘ )− sin 2 ( 28 ∘ )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>2</mn><msup/></mrow></annotation-xml></semantics></math> cos 2 ( 37 ∘ )−1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mo stretchy="false">(</mo><msup/></mrow></annotation-xml></semantics></math> 74 ∘ )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>1</mn><mo>−</mo><mn>2</mn><msup/></mrow></annotation-xml></semantics></math>  sin 2 ( 17 ∘ )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>cos</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 (9x)− sin 2 (9x)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mo stretchy="false">(</mo><mn>18</mn><mi>x</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics></math>

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>4</mn><mtext> </mtext><mi>sin</mi><mo stretchy="false">(</mo><mn>8</mn><mi>x</mi><mo stretchy="false">)</mo><mtext> </mtext><mi>cos</mi><mo stretchy="false">(</mo><mn>8</mn><mi>x</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics></math>

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>6</mn><mtext> </mtext><mi>sin</mi><mo stretchy="false">(</mo><mn>5</mn><mi>x</mi><mo stretchy="false">)</mo><mtext> </mtext><mi>cos</mi><mo stretchy="false">(</mo><mn>5</mn><mi>x</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics></math>

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>3</mn><mi>sin</mi><mo stretchy="false">(</mo><mn>10</mn><mi>x</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics></math>

For the following exercises, prove the identity given.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></annotation-xml></semantics></math> sin t−cos t ) 2 =1−sin( 2t )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2x )=−2 sin( −x ) cos( −x )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mn>2</mn><mtext> </mtext><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> −x )cos( −x )=−2(−sin( x )cos( x ))=sin( 2x )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cot</mi><mtext> </mtext><mi>x</mi><mo>−</mo><mi>tan</mi><mtext> </mtext><mi>x</mi><mo>=</mo><mn>2</mn><mtext> </mtext><mi>cot</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2x )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></mfrac></mrow></annotation-xml></semantics></math> 2θ ) 1+cos( 2θ ) tan 2 θ=tan θ

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><mfrac><mrow><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></mfrac></mtd></mtr></mtable></annotation-xml></semantics></math> 2θ ) 1+cos( 2θ ) tan 2 θ= 2sin( θ )cos( θ ) 1+ cos 2 θ− sin 2 θ tan 2 θ= 2sin( θ )cos( θ ) 2 cos 2 θ tan 2 θ= sin( θ ) cos θ tan2 θ= cot( θ ) tan 2 θ=tan θ

For the following exercises, rewrite the expression with an exponent no higher than 1.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>cos</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 (5x)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>cos</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 (6x)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>1</mn><mo>+</mo><mi>cos</mi><mo stretchy="false">(</mo><mn>12</mn><mi>x</mi><mo stretchy="false">)</mo></mrow></mfrac></mrow></annotation-xml></semantics></math> 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>sin</mi></mrow></msup></mrow></annotation-xml></semantics></math> 4 (8x)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>sin</mi></mrow></msup></mrow></annotation-xml></semantics></math> 4 (3x)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>3</mn><mo>+</mo><mi>cos</mi><mo stretchy="false">(</mo><mn>12</mn><mi>x</mi><mo stretchy="false">)</mo><mo>−</mo><mn>4</mn><mi>cos</mi><mo stretchy="false">(</mo><mn>6</mn><mi>x</mi><mo stretchy="false">)</mo></mrow></mfrac></mrow></annotation-xml></semantics></math> 8

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>cos</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 x  sin 4 x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>cos</mi></mrow></msup></mrow></annotation-xml></semantics></math> 4 x  sin 2 x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>2</mn><mo>+</mo><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>x</mi><mo stretchy="false">)</mo><mo>−</mo><mn>2</mn><mi>cos</mi><mo stretchy="false">(</mo><mn>4</mn><mi>x</mi><mo stretchy="false">)</mo><mo>−</mo><mi>cos</mi><mo stretchy="false">(</mo><mn>6</mn><mi>x</mi><mo stretchy="false">)</mo></mrow></mfrac></mrow></annotation-xml></semantics></math> 32

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>tan</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 x  sin 2 x

Technology

For the following exercises, reduce the equations to powers of one, and then check the answer graphically.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>tan</mi></mrow></msup></mrow></annotation-xml></semantics></math> 4 x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>3</mn><mo>+</mo><mi>cos</mi><mo stretchy="false">(</mo><mn>4</mn><mi>x</mi><mo stretchy="false">)</mo><mo>−</mo><mn>4</mn><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>x</mi><mo stretchy="false">)</mo></mrow></mfrac></mrow></annotation-xml></semantics></math> 3+cos(4x)+4cos(2x)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>sin</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 (2x)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>sin</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 x  cos 2 x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>1</mn><mo>−</mo><mi>cos</mi><mo stretchy="false">(</mo><mn>4</mn><mi>x</mi><mo stretchy="false">)</mo></mrow></mfrac></mrow></annotation-xml></semantics></math> 8

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>tan</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 x sin x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>tan</mi></mrow></msup></mrow></annotation-xml></semantics></math> 4 x  cos 2 x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>3</mn><mo>+</mo><mi>cos</mi><mo stretchy="false">(</mo><mn>4</mn><mi>x</mi><mo stretchy="false">)</mo><mo>−</mo><mn>4</mn><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>x</mi><mo stretchy="false">)</mo></mrow></mfrac></mrow></annotation-xml></semantics></math> 4(cos(2x)+1)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>cos</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 x sin( 2x )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>cos</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 ( 2x )sin x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mrow><mo>(</mo></mrow></mrow></mfrac></mrow></annotation-xml></semantics></math> 1+cos( 4x ) )sin x 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mi>tan</mi></mrow></msup></mrow></annotation-xml></semantics></math> 2 ( x 2 ) sin x

For the following exercises, algebraically find an equivalent function, only in terms of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>sin</mi><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>and/or<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>cos</mi><mtext> </mtext><mi>x</mi><mo>,</mo></mrow></annotation-xml></semantics></math>and then check the answer by graphing both equations.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>sin</mi><mo stretchy="false">(</mo><mn>4</mn><mi>x</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics></math>

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>4</mn><mi>sin</mi><mtext> </mtext><mi>x</mi><mi>cos</mi><mtext> </mtext><mi>x</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> cos 2 x− sin 2 x )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mo stretchy="false">(</mo><mn>4</mn><mi>x</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics></math>

Extensions

For the following exercises, prove the identities.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 2x )= 2 tan x 1+ tan 2 x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>2</mn><mi>tan</mi><mtext> </mtext><mi>x</mi></mrow></mfrac></mrow></annotation-xml></semantics></math> 1+ tan 2 x = 2sin x cos x 1+ sin 2 x cos 2 x = 2sin x cos x cos 2 x+ sin 2 x cos 2 x =

 
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>2</mn><mi>sin</mi><mspace width="0.2em"/><mi>x</mi></mrow></mfrac></mrow></annotation-xml></semantics></math> cosx . cos 2 x 1 =2sinxcosx=sin(2x)

 

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>α</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 1− tan 2 α 1+ tan 2 α

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>tan</mi><mo stretchy="false">(</mo><mn>2</mn><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 2 sin xcos x 2 cos 2 x−1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>2</mn><mi>sin</mi><mtext> </mtext><mi>x</mi><mi>cos</mi><mtext> </mtext><mi>x</mi></mrow></mfrac></mrow></annotation-xml></semantics></math> 2 cos 2 x−1 = sin(2x) cos(2x) =tan(2x)

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mrow><mo>(</mo></mrow></mrow></msup></mrow></annotation-xml></semantics></math> sin 2 x−1 ) 2 =cos( 2x )+ sin 4 x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 3x )=3 sin x  cos 2 x− sin 3 x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>sin</mi><mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mn>2</mn><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mi>sin</mi><mtext> </mtext><mi>x</mi><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>x</mi><mo stretchy="false">)</mo><mo>+</mo><mi>sin</mi><mo stretchy="false">(</mo><mn>2</mn><mi>x</mi><mo stretchy="false">)</mo><mi>cos</mi><mtext> </mtext><mi>x</mi></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math>                             =sin x( cos 2 x− sin 2 x)+2sin xcos xcos x                             =sin x cos 2 x−sin 3 x+2sin x cos 2 x                             =3sin x cos 2 x− sin 3 x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 3x )= cos 3 x−3 sin 2 xcos x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>1</mn><mo>+</mo><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></mfrac></mrow></annotation-xml></semantics></math> 2t ) sin( 2t )−cos t = 2 cos t 2 sin t−1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mfrac><mrow><mn>1</mn><mo>+</mo><mi>cos</mi><mo stretchy="false">(</mo><mn>2</mn><mi>t</mi><mo stretchy="false">)</mo></mrow></mfrac></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> sin(2t)−cost = 1+2 cos 2 t−1 2sintcost−cost                           = 2 cos 2 t cost(2sint−1)                           = 2cost 2sint−1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>sin</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 16x )=16 sin x cos x cos( 2x )cos( 4x )cos( 8x )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>cos</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 16x )=( cos 2 ( 4x )− sin 2 ( 4x )−sin( 8x ) )( cos 2 ( 4x )− sin 2 ( 4x )+sin( 8x ) )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mrow><mo>(</mo></mrow></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> cos 2 (4x)− sin 2 (4x)−sin(8x))( cos 2 (4x)− sin 2 (4x)+sin(8x) )=                                                                                                  =( cos(8x)−sin(8x))(cos(8x)+sin(8x) )                                                                                                  = cos 2 (8x)− sin 2 (8x)                                                                                                  =cos(16x)

Glossary

double-angle formulas
identities derived from the sum formulas for sine, cosine, and tangent in which the angles are equal
half-angle formulas
identities derived from the reduction formulas and used to determine half-angle values of trigonometric functions
reduction formulas
identities derived from the double-angle formulas and used to reduce the power of a trigonometric function