1.3: Trigonometric Functions

Learning Objectives

• Convert angle measures between degrees and radians.
• Recognize the triangular and circular definitions of the basic trigonometric functions.
• Verify the fundamental trigonometric identities.
• Solve trigonometric equations.

Trigonometric functions are used to model many phenomena, including sound waves, vibrations of strings, alternating electrical current, and the motion of pendulums. In fact, almost any repetitive, or cyclical, motion can be modeled by some combination of trigonometric functions. In this section, we define the six basic trigonometric functions and look at some of the main identities involving these functions.

Radian Measure

To use trigonometric functions, we first must understand how to measure the angles. Although we can use both radians and degrees, radians are a more natural measurement because they are related directly to the unit circle, a circle with radius 1. The radian measure of an angle is defined as follows. Given an angle \(θ\), let \(s\) be the length of the corresponding arc on the unit circle (Figure \(\PageIndex{1}\)). We say the angle corresponding to the arc of length 1 has radian measure 1.

Figure \(\PageIndex{1}\): The radian measure of an angle \(θ\) is the arc length \(s\) of the associated arc on the unit circle.

Since an angle of \(360°\) corresponds to the circumference of a circle, or an arc of length \(2π\), we conclude that an angle with a degree measure of \(360°\) has a radian measure of \(2π\). Similarly, we see that \(180°\) is equivalent to \(\pi\) radians. Table \(\PageIndex{1}\) shows the relationship between common degree and radian values.

Table \(\PageIndex{1}\): Common Angles Expressed in Degrees and Radians

Degrees	Radians	Degrees	Radians
0	0	120	\(2π/3\)
30	\(π/6\)	135	\(3π/4\)
45	\(π/4\)	150	\(5π/6\)
60	\(π/3\)	180	\(π\)
90	\(π/2\)

Converting between Radians and Degrees

1. Express \(225°\) using radians.
2. Express \(5π/3\) rad using degrees.

Solution

Use the fact that \(180\)° is equivalent to \(\pi\) radians as a conversion factor (Table \(\PageIndex{1}\)):

\[1=\dfrac{π \,\mathrm{rad}}{180°}=\dfrac{180°}{π \,\mathrm{rad}}. \nonumber \]

1. \(225°=225°⋅\left(\dfrac{π}{180°}\right)=\left(\dfrac{5π}{4}\right)\) rad
2. \(\dfrac{5π}{3}\) rad = \(\dfrac{5π}{3}\)⋅\(\dfrac{180°}{π}\)=\(300\)°

Exercise \(\PageIndex{1}\)

1. Express \(210°\) using radians.
2. Express \(11π/6\) rad using degrees.

Hint

\(π\) radians is equal to 180°

Answer

1. \(7π/6\)
2. 330°

The Six Basic Trigonometric Functions

Trigonometric functions allow us to use angle measures, in radians or degrees, to find the coordinates of a point on any circle—not only on a unit circle—or to find an angle given a point on a circle. They also define the relationship between the sides and angles of a triangle.

To define the trigonometric functions, first consider the unit circle centered at the origin and a point \(P=(x,y)\) on the unit circle. Let \(θ\) be an angle with an initial side that lies along the positive \(x\)-axis and with a terminal side that is the line segment \(OP\). An angle in this position is said to be in standard position (Figure \(\PageIndex{2}\)). We can then define the values of the six trigonometric functions for \(θ\) in terms of the coordinates \(x\) and \(y\).

Figure \(\PageIndex{2}\): The angle \(θ\) is in standard position. The values of the trigonometric functions for \(θ\) are defined in terms of the coordinates \(x\) and \(y\).

Definition: Trigonometric functions

Let \(P=(x,y)\) be a point on the unit circle centered at the origin \(O\). Let \(θ\) be an angle with an initial side along the positive \(x\)-axis and a terminal side given by the line segment \(OP\). The trigonometric functions are then defined as

\(\sin θ=y\)	\(\csc θ=\dfrac{1}{y}\)
\(\cos θ=x\)	\(\sec θ=\dfrac{1}{x}\)
\(\tan θ=\dfrac{y}{x}\)	\(\cot θ=\dfrac{x}{y}\)

If \(x=0, \sec θ\) and \(\tan θ\) are undefined. If \(y=0\), then \(\cot θ\) and \(\csc θ\) are undefined.

We can see that for a point \(P=(x,y)\) on a circle of radius \(r\) with a corresponding angle \(θ\), the coordinates \(x\) and \(y\) satisfy

\[\begin{align} \cos θ &=\dfrac{x}{r} \\ x &=r\cos θ \end{align} \nonumber \]

and

\[\begin{align} \sin θ &=\dfrac{y}{r} \\ y &=r\sin θ. \end{align} \nonumber \]

The values of the other trigonometric functions can be expressed in terms of \(x,y\), and \(r\) (Figure \(\PageIndex{3}\)).

Figure \(\PageIndex{3}\): For a point \(P=(x,y)\) on a circle of radius \(r\), the coordinates \(x\) and \(y\) satisfy \(x=r\cos θ\) and \(y=r\sin θ\).

Table \(\PageIndex{2}\) shows the values of sine and cosine at the major angles in the first quadrant. From this table, we can determine the values of sine and cosine at the corresponding angles in the other quadrants. The values of the other trigonometric functions are calculated easily from the values of \(\sin θ\) and \(\cos θ.\)

Table \(\PageIndex{2}\): Values of \(\sin θ\) and \(\cos θ\) at Major Angles \(θ\) in the First Quadrant

\(θ\)θ	\(\sin θ\)	\(\cos θ\)
0	0	1
\(\dfrac{π}{6}\)	\(\dfrac{1}{2}\)	\(\dfrac{\sqrt{3}}{2}\)
\(\dfrac{π}{4}\)	\(\dfrac{\sqrt{2}}{2}\)	\(\dfrac{\sqrt{2}}{2}\)
\(\dfrac{π}{3}\)	\(\dfrac{\sqrt{3}}{2}\)	\(\dfrac{1}{2}\)
\(\dfrac{π}{2}\)	1	0

Example \(\PageIndex{2}\): Evaluating Trigonometric Functions

Evaluate each of the following expressions.

1. \(\sin \left(\dfrac{2π}{3} \right)\)
2. \(\cos \left(−\dfrac{5π}{6} \right)\)
3. \(\tan \left(\dfrac{15π}{4}\right)\)

Solution:

a) On the unit circle, the angle \(θ=\dfrac{2π}{3}\) corresponds to the point \(\left(−\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)\). Therefore,

\[ \sin \left(\dfrac{2π}{3}\right)=y=\left(\dfrac{\sqrt{3}}{2}\right). \nonumber \]

b) An angle \(θ=−\dfrac{5π}{6}\) corresponds to a revolution in the negative direction, as shown. Therefore,

\[\cos \left(−\dfrac{5π}{6}\right)=x=−\dfrac{\sqrt{3}}{2}. \nonumber \]

c) An angle \(θ\)=\(\dfrac{15π}{4}\)=\(2π\)+\(\dfrac{7π}{4}\). Therefore, this angle corresponds to more than one revolution, as shown. Knowing the fact that an angle of \(\dfrac{7π}{4}\) corresponds to the point \((\dfrac{\sqrt{2}}{2},-\dfrac{\sqrt{2}}{2})\), we can conclude that

\[\tan \left(\dfrac{15π}{4}\right)=\dfrac{y}{x}=−1. \nonumber \]

Exercise \(\PageIndex{2}\)

Evaluate \(\cos(3π/4)\) and \(\sin(−π/6)\).

Hint

Look at angles on the unit circle.

Answer

\[\cos(3π/4) = −\sqrt{2}/2\nonumber \]

\[ \sin(−π/6) =−1/2 \nonumber \]

As mentioned earlier, the ratios of the side lengths of a right triangle can be expressed in terms of the trigonometric functions evaluated at either of the acute angles of the triangle. Let \(θ\) be one of the acute angles. Let \(A\) be the length of the adjacent leg, \(O\) be the length of the opposite leg, and \(H\) be the length of the hypotenuse. By inscribing the triangle into a circle of radius \(H\), as shown in Figure \(\PageIndex{4}\), we see that \(A,H\), and \(O\) satisfy the following relationships with \(θ\):

\(\sin θ=\dfrac{O}{H}\)	\(\csc θ=\dfrac{H}{O}\)
\(\cos θ=\dfrac{A}{H}\)	\(\sec θ=\dfrac{H}{A}\)
\(\tan θ=\dfrac{O}{A}\)	\(\cot θ=\dfrac{A}{O}\)

Figure \(\PageIndex{4}\): By inscribing a right triangle in a circle, we can express the ratios of the side lengths in terms of the trigonometric functions evaluated at \(θ\).

Example \(\PageIndex{3}\): Constructing a Wooden Ramp

A wooden ramp is to be built with one end on the ground and the other end at the top of a short staircase. If the top of the staircase is \(4\) ft from the ground and the angle between the ground and the ramp is to be \(10\)°, how long does the ramp need to be?

Solution

Let \(x\) denote the length of the ramp. In the following image, we see that \(x\) needs to satisfy the equation \(\sin(10°)=4/x\). Solving this equation for \(x\), we see that \(x=4/\sin(10°)\)≈\(23.035\) ft.

Exercise \(\PageIndex{3}\)

A house painter wants to lean a \(20\)-ft ladder against a house. If the angle between the base of the ladder and the ground is to be \(60\)°, how far from the house should she place the base of the ladder?

Hint

Draw a right triangle with hypotenuse 20.

Answer

10 ft

Verifying the Fundamental Trigonometric Identities

Identities enable us to simplify complicated expressions. They are the basic tools of trigonometry used in solving trigonometric equations, just as factoring, finding common denominators, and using special formulas are the basic tools of solving algebraic equations. In fact, we use algebraic techniques constantly to simplify trigonometric expressions. Basic properties and formulas of algebra, such as the difference of squares formula and the perfect squares formula, will simplify the work involved with trigonometric expressions and equations. We already know that all of the trigonometric functions are related because they all are defined in terms of the unit circle. Consequently, any trigonometric identity can be written in many ways.

To verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation. Sometimes we have to factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desired result. In this first section, we will work with the fundamental identities: the Pythagorean identities, the even-odd identities, the reciprocal identities, and the quotient identities.

We will begin with the Pythagorean identities (Table \(\PageIndex{1}\)), which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identifies, but now we will also use additional identities.

Table \(\PageIndex{1}\): Pythagorean Identities

\({\sin}^2 \theta+{\cos}^2 \theta=1\)

\(1+{\cot}^2 \theta={\csc}^2 \theta\)

\(1+{\tan}^2 \theta={\sec}^2 \theta\)

The second and third identities can be obtained by manipulating the first. The identity \(1+{\cot}^2 \theta={\csc}^2 \theta\) is found by rewriting the left side of the equation in terms of sine and cosine.

Prove: \(1+{\cot}^2 \theta={\csc}^2 \theta\)

\[\begin{align*} 1+{\cot}^2 \theta&= (1+\dfrac{{\cos}^2}{{\sin}^2})\qquad \text{Rewrite the left side}\\ &= \left(\dfrac{{\sin}^2}{{\sin}^2}\right)+\left (\dfrac{{\cos}^2}{{\sin}^2}\right)\qquad \text{Write both terms with the common denominator}\\ &= \dfrac{{\sin}^2+{\cos}^2}{{\sin}^2}\\ &= \dfrac{1}{{\sin}^2}\\ &= {\csc}^2 \end{align*}\]

Similarly,\(1+{\tan}^2 \theta={\sec}^2 \theta\)can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives

\[\begin{align*} 1+{\tan}^2 \theta&= 1+{\left(\dfrac{\sin \theta}{\cos \theta}\right )}^2\qquad \text{Rewrite left side}\\ &= {\left (\dfrac{\cos \theta}{\cos \theta}\right )}^2+{\left (\dfrac{\sin \theta}{\cos \theta}\right)}^2\qquad \text{Write both terms with the common denominator}\\ &= \dfrac{{\cos}^2 \theta+{\sin}^2 \theta}{{\cos}^2 \theta}\\ &= \dfrac{1}{{\cos}^2 \theta}\\ &= {\sec}^2 \theta \end{align*}\]

Recall that we determined which trigonometric functions are odd and which are even. The next set of fundamental identities is the set of even-odd identities. The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle (Table \(\PageIndex{2}\)).

Table \(\PageIndex{2}\): Even-Odd Identities

\(\tan(−\theta)=−\tan \theta\)	\(\sin(−\theta)=−\sin \theta\)	\(\cos(−\theta)=\cos \theta\)
\(\cot(−\theta)=−\cot \theta\)	\(\csc(−\theta)=−\csc \theta\)	\(\sec(−\theta)=\sec \theta\)

Recall that an odd function is one in which \(f(−x)= −f(x)\) for all \(x\) in the domain off. f. The sine function is an odd function because \(\sin(−\theta)=−\sin \theta\). The graph of an odd function is symmetric about the origin. For example, consider corresponding inputs of \(\dfrac{\pi}{2}\) and \(−\dfrac{\pi}{2}\). The output of \(\sin\left (\dfrac{\pi}{2}\right )\) is opposite the output of \(\sin \left (−\dfrac{\pi}{2}\right )\). Thus,

\[\begin{align*} \sin\left (\dfrac{\pi}{2}\right)&=1 \\[4pt] \sin\left (-\dfrac{\pi}{2}\right) &=-\sin\left (\dfrac{\pi}{2}\right) \\[4pt] &=-1 \end{align*}\]

This is shown in Figure \(\PageIndex{2}\).

Figure \(\PageIndex{2}\): Graph of \(y=\sin \theta\)

Recall that an even function is one in which

\(f(−x)=f(x)\) for all \(x\) in the domain of \(f\)

The graph of an even function is symmetric about the y-axis. The cosine function is an even function because \(\cos(−\theta)=\cos \theta\). For example, consider corresponding inputs \(\dfrac{\pi}{4}\) and \(−\dfrac{\pi}{4}\). The output of \(\cos\left (\dfrac{\pi}{4}\right)\) is the same as the output of \(\cos\left (−\dfrac{\pi}{4}\right)\). Thus,

\[\begin{align*} \cos\left (−\dfrac{\pi}{4}\right ) &=\cos\left (\dfrac{\pi}{4}\right) \\[4pt] &≈0.707 \end{align*}\]

See Figure \(\PageIndex{3}\).

Figure \(\PageIndex{3}\): Graph of \(y=\cos \theta\)

For all \(\theta\) in the domain of the sine and cosine functions, respectively, we can state the following:

• Since \(\sin(−\theta)=−\sin \theta\),sine is an odd function.
• Since \(\cos(−\theta)=\cos \theta\),cosine is an even function.

The other even-odd identities follow from the even and odd nature of the sine and cosine functions. For example, consider the tangent identity,\(\tan(−\theta)=−\tan \theta\). We can interpret the tangent of a negative angle as

\[\tan (−\theta)=\dfrac{\sin (−\theta)}{\cos (−\theta)}=\dfrac{−\sin \theta}{\cos \theta}=−\tan \theta. \nonumber\]

Tangent is therefore an odd function, which means that \(\tan(−\theta)=−\tan(\theta)\) for all \(\theta\) in the domain of the tangent function.

The cotangent identity, \(\cot(−\theta)=−\cot \theta\),also follows from the sine and cosine identities. We can interpret the cotangent of a negative angle as

\[\cot(−\theta)=\dfrac{\cos(−\theta)}{\sin(−\theta)}=\dfrac{\cos \theta}{−\sin \theta}=−\cot \theta.\nonumber\]

Cotangent is therefore an odd function, which means that \(\cot(−\theta)=−\cot(\theta)\) for all \(\theta\) in the domain of the cotangent function.

The cosecant function is the reciprocal of the sine function, which means that the cosecant of a negative angle will be interpreted as

\[\csc(−\theta)=\dfrac{1}{\sin(−\theta)}=\dfrac{1}{−\sin \theta}=−\csc \theta. \nonumber\]

The cosecant function is therefore odd.

Finally, the secant function is the reciprocal of the cosine function, and the secant of a negative angle is interpreted as

\[\sec(−\theta)=\dfrac{1}{\cos(−\theta)}=\dfrac{1}{\cos \theta}=\sec \theta. \nonumber\]

The secant function is therefore even.

To sum up, only two of the trigonometric functions, cosine and secant, are even. The other four functions are odd, verifying the even-odd identities.

The next set of fundamental identities is the set of reciprocal identities, which, as their name implies, relate trigonometric functions that are reciprocals of each other. (Table \(\PageIndex{3}\)). Recall that we first encountered these identities when defining trigonometric functions from right angles in Right Angle Trigonometry.

Table \(\PageIndex{3}\): Reciprocal Identities

\(\sin \theta=\dfrac{1}{\csc \theta}\)	\(\csc \theta=\dfrac{1}{\sin \theta}\)
\(\cos \theta = \dfrac{1}{\sec \theta}\)	\(\sec \theta=\dfrac{1}{\cos \theta}\)
\(\tan \theta=\dfrac{1}{\cot \theta}\)	\(\cot \theta=\dfrac{1}{\tan \theta}\)

The final set of identities is the set of quotient identities, which define relationships among certain trigonometric functions and can be very helpful in verifying other identities (Table \(\PageIndex{4}\)).

Table \(\PageIndex{4}\): Quotient Identities

\(\tan \theta=\dfrac{\sin \theta}{\cos \theta}\)	\(\cot \theta=\dfrac{\cos \theta}{\sin \theta}\)

The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions.

SUMMARIZING TRIGONOMETRIC IDENTITIES

The Pythagorean identities are based on the properties of a right triangle.

\[{\cos}^2 \theta+{\sin}^2 \theta=1\]

\[1+{\cot}^2 \theta={\csc}^2 \theta\]

\[1+{\tan}^2 \theta={\sec}^2 \theta\]

The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle.

\[\tan(−\theta)=−\tan \theta\]

\[\cot(−\theta)=−\cot \theta\]

\[\sin(−\theta)=−\sin \theta\]

\[\csc(−\theta)=−\csc \theta\]

\[\cos(−\theta)=\cos \theta\]

\[\sec(−\theta)=\sec \theta\]

The reciprocal identities define reciprocals of the trigonometric functions.

\[\sin \theta=\dfrac{1}{\csc \theta}\]

\[\cos \theta=\dfrac{1}{\sec \theta}\]

\[\tan \theta=\dfrac{1}{\cot \theta}\]

\[\csc \theta=\dfrac{1}{\sin \theta}\]

\[\sec \theta=\dfrac{1}{\cos \theta}\]

\[\cot \theta=\dfrac{1}{\tan \theta}\]

The quotient identities define the relationship among the trigonometric functions.

\[\tan \theta=\dfrac{\sin \theta}{\cos \theta}\]

\[\cot \theta=\dfrac{\cos \theta}{\sin \theta}\]

How to: Given a trigonometric identity, verify that it is true.

1. Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build.
2. Look for opportunities to factor expressions, square a binomial, or add fractions.
3. Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.
4. If these steps do not yield the desired result, try converting all terms to sines and cosines.

Example \(\PageIndex{2}\): Verifying a Trigonometric Identity

Verify \(\tan \theta \cos \theta=\sin \theta\).

Solution

We will start on the left side, as it is the more complicated side:

\[ \begin{align*} \tan \theta \cos \theta &=\left(\dfrac{\sin \theta}{\cos \theta}\right)\cos \theta \\[4pt] &=\sin \theta \end{align*}\]

Analysis

This identity was fairly simple to verify, as it only required writing \(\tan \theta\) in terms of \(\sin \theta\) and \(\cos \theta\).

Exercise \(\PageIndex{1}\)

Verify the identity \(\csc \theta \cos \theta \tan \theta=1\).

Answer

\[ \begin{align*} \csc \theta \cos \theta \tan \theta=\left(\dfrac{1}{\sin \theta}\right)\cos \theta\left(\dfrac{\sin \theta}{\cos \theta}\right) \\[4pt] & =\dfrac{\cos \theta}{\sin \theta}(\dfrac{\sin \theta}{\cos \theta}) \\[4pt] & =\dfrac{\sin \theta \cos \theta}{\sin \theta \cos \theta} \\[4pt] &=1 \end{align*}\]

Example \(\PageIndex{3A}\): Verifying the Equivalency Using the Even-Odd Identities

Verify the following equivalency using the even-odd identities:

\((1+\sin x)[1+\sin(−x)]={\cos}^2 x\)

Solution

Working on the left side of the equation, we have

\( (1+\sin x)[1+\sin(−x)]=(1+\sin x)(1-\sin x)\)

Since

\[\begin{align*} \sin(-x)&= -\sin x \\ [5pt] &=1-{\sin}^2 x\qquad \text{Difference of squares} \\ [5pt] &={\cos}^2 x \\ {\cos}^2 x&= 1-{\sin}^2 x \\ \end{align*}\]

Example \(\PageIndex{3B}\): Verifying a Trigonometric Identity Involving \({\sec}^2 \theta\)

Verify the identity \(\dfrac{{\sec}^2 \theta−1}{{\sec}^2 \theta}={\sin}^2 \theta\)

Solution

As the left side is more complicated, let’s begin there.

\[\begin{align*}
\dfrac{{\sec}^2 \theta-1}{{\sec}^2 \theta}&= \dfrac{({\tan}^2 \theta +1)-1}{{\sec}^2 \theta}\\
{\sec}^2 \theta&= {\tan}^2 \theta +1\\
&= \dfrac{{\tan}^2 \theta}{{\sec}^2 \theta}\\
&= {\tan}^2 \theta\left (\dfrac{1}{{\sec}^2 \theta}\right )\\
&= {\tan}^2 \theta \left ({\cos}^2 \theta\right )\\
{\cos}^2 \theta&= \dfrac{1}{{\sec}^2 \theta}\\
&= \left (\dfrac{{\sin}^2 \theta}{{\cos}^2 \theta}\right )\\
{\tan}^2 \theta&= \dfrac{{\sin}^2 \theta}{{\cos}^2 \theta}\\
&= {\sin}^2 \theta
\end{align*}\]

There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.

\[\begin{align*} \dfrac{{\sec}^2 \theta-1}{{\sec}^2 \theta}&= \dfrac{{\sec}^2 \theta}{{\sec}^2 \theta}-\dfrac{1}{{\sec}^2 \theta}\\ &= 1-{\cos}^2 \theta\\ &= {\sin}^2 \theta \end{align*}\]

Analysis

In the first method, we used the identity \({\sec}^2 \theta={\tan}^2 \theta+1\) and continued to simplify. In the second method, we split the fraction, putting both terms in the numerator over the common denominator. This problem illustrates that there are multiple ways we can verify an identity. Employing some creativity can sometimes simplify a procedure. As long as the substitutions are correct, the answer will be the same.

Exercise \(\PageIndex{2}\)

Show that \(\dfrac{\cot \theta}{\csc \theta}=\cos \theta\).

Answer

\[\begin{align*} \dfrac{\cot \theta}{\csc \theta}&= \dfrac{\tfrac{\cos \theta}{\sin \theta}}{\dfrac{1}{\sin \theta}}\\ &= \dfrac{\cos \theta}{\sin \theta}\cdot \dfrac{\sin \theta}{1}\\ &= \cos \theta \end{align*}\]

Example \(\PageIndex{4}\): Creating and Verifying an Identity

Create an identity for the expression \(2 \tan \theta \sec \theta\) by rewriting strictly in terms of sine.

Solution

There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:

\[\begin{align*} 2 \tan \theta \sec \theta&= 2\left (\dfrac{\sin \theta}{\cos \theta}\right )\left(\dfrac{1}{\cos \theta}\right )\\ &= \dfrac{2\sin \theta}{{\cos}^2 \theta}\\ &= \dfrac{2\sin \theta}{1-{\sin}^2 \theta}\qquad \text{Substitute } 1-{\sin}^2 \theta \text{ for } {\cos}^2 \theta \end{align*}\]

Thus,

\(2 \tan \theta \sec \theta=\dfrac{2 \sin \theta}{1−{\sin}^2 \theta}\)

Example \(\PageIndex{5}\): Verifying an Identity Using Algebra and Even/Odd Identities

Verify the identity:

\(\dfrac{{\sin}^2(−\theta)−{\cos}^2(−\theta)}{\sin(−\theta)−\cos(−\theta)}=\cos \theta−\sin \theta\)

Solution

Let’s start with the left side and simplify:

\[\begin{align*} \dfrac{{\sin}^2(-\theta)-{\cos}^2(-\theta)}{\sin(-\theta)-\cos(-\theta)}&= \dfrac{{[\sin(-\theta)]}^2-{[\cos(-\theta)]}^2}{\sin(-\theta)-\cos(-\theta)}\\ &= \dfrac{{(-\sin \theta)}^2-{(\cos \theta)}^2}{-\sin \theta -\cos \theta} \;\; \; , \sin(-x) = -\sin\space x\text { and } \cos(-x)=\cos \space x\\ &= \dfrac{{(\sin \theta)}^2-{(\cos \theta)}^2}{-\sin \theta -\cos \theta}\qquad \text{Difference of squares}\\ &= \dfrac{(\sin \theta-\cos \theta)(\sin \theta+\cos \theta)}{-(\sin \theta+\cos \theta)}\\ &= \cos \theta-\sin \theta \end{align*}\]

Exercise \(\PageIndex{3}\)

Verify the identity \(\dfrac{{\sin}^2 \theta−1}{\tan \theta \sin \theta−\tan \theta}=\dfrac{\sin \theta+1}{\tan \theta}\).

Answer

\[\begin{align*} \dfrac{{\sin}^2 \theta-1}{\tan \theta \sin \theta-\tan \theta}&= \dfrac{(\sin \theta +1)(\sin \theta -1)}{\tan \theta(\sin \theta -1)}\\ &= \dfrac{\sin \theta+1}{\tan \theta} \end{align*}\]

Example \(\PageIndex{6}\): Verifying an Identity Involving Cosines and Cotangents

Verify the identity: \((1−{\cos}^2 x)(1+{\cot}^2 x)=1\).

Solution

\[\begin{align*} (1-{\cos}^2 x)(1+{\cot}^2 x)&= (1-{\cos}^2 x)\left(1+\dfrac{{\cos}^2 x}{{\sin}^2 x}\right)\\ &= (1-{\cos}^2 x)\left(\dfrac{{\sin}^2 x}{{\sin}^2 x}+\dfrac{{\cos}^2 x}{{\sin}^2 x}\right )\qquad \text{Find the common denominator}\\ &= (1-{\cos}^2 x)\left(\dfrac{{\sin}^2 x +{\cos}^2 x}{{\sin}^2 x}\right)\\ &= ({\sin}^2 x)\left (\dfrac{1}{{\sin}^2 x}\right )\\ &= 1 \end{align*}\]

Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see [link]). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is \(\pi\),not \(2\pi\). Further, the domain of tangent is all real numbers with the exception of odd integer multiples of \(\dfrac{\pi}{2}\),unless, of course, a problem places its own restrictions on the domain.

Example \(\PageIndex{7A}\): Solving a Problem Involving a Single Trigonometric Function

Solve the problem exactly: \(2 {\sin}^2 \theta−1=0\), \(0≤\theta<2\pi\).

Solution

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate \(\sin \theta\). Then we will find the angles.

\[\begin{align*}
2 {\sin}^2 \theta-1&= 0\\
2 {\sin}^2 \theta&= 1\\
{\sin}^2 \theta&= \dfrac{1}{2}\\
\sqrt{ {\sin}^2 \theta }&= \pm \sqrt{ \dfrac{1}{2} }\\
\sin \theta&= \pm \dfrac{1}{\sqrt{2}}\\
&= \pm \dfrac{\sqrt{2}}{2}\\
\theta&= \dfrac{\pi}{4}, \space \dfrac{3\pi}{4},\space \dfrac{5\pi}{4}, \space \dfrac{7\pi}{4}
\end{align*}\]

Example \(\PageIndex{7B}\): Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: \(\csc \theta=−2\), \(0≤\theta<4\pi\).

Solution

We want all values of \(\theta\) for which \(\csc \theta=−2\) over the interval \(0≤\theta<4\pi\).

\[\begin{align*} \csc \theta&= -2\\ \dfrac{1}{\sin \theta}&= -2\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\space \dfrac{11\pi}{6},\space \dfrac{19\pi}{6}, \space \dfrac{23\pi}{6} \end{align*}\]

Analysis

As \(\sin \theta=−\dfrac{1}{2}\), notice that all four solutions are in the third and fourth quadrants.

Example \(\PageIndex{7C}\): Solving an Equation Involving Tangent

Solve the equation exactly: \(\tan\left(\theta−\dfrac{\pi}{2}\right)=1\), \(0≤\theta<2\pi\).

Solution

Recall that the tangent function has a period of \(\pi\). On the interval \([ 0,\pi )\),and at the angle of \(\dfrac{\pi}{4}\),the tangent has a value of \(1\). However, the angle we want is \(\left(\theta−\dfrac{\pi}{2}\right)\). Thus, if \(\tan\left(\dfrac{\pi}{4}\right)=1\),then

\[\begin{align*} \theta-\dfrac{\pi}{2}&= \dfrac{\pi}{4}\\ \theta&= \dfrac{3\pi}{4} \pm k\pi \end{align*}\]

Over the interval \([ 0,2\pi )\),we have two solutions:

\(\theta=\dfrac{3\pi}{4}\) and \(\theta=\dfrac{3\pi}{4}+\pi=\dfrac{7\pi}{4}\)

Exercise \(\PageIndex{4}\)

Find all solutions for \(\tan x=\sqrt{3}\).

Answer

\(\dfrac{\pi}{3}\pm \pi k\)

Example \(\PageIndex{8}\): Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation \(2(\tan x+3)=5+\tan x\), \(0≤x<2\pi\).

Solution

We can solve this equation using only algebra. Isolate the expression \(\tan x\) on the left side of the equals sign.

\[\begin{align*} 2(\tan x)+2(3)&= 5+\tan x\\ 2\tan x+6&= 5+\tan x\\ 2\tan x-\tan x&= 5-6\\ \tan x&= -1 \end{align*}\]

There are two angles on the unit circle that have a tangent value of \(−1\): \(\theta=\dfrac{3\pi}{4}\) and \(\theta=\dfrac{7\pi}{4}\).

Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as \(x\) or \(u\). If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

Example \(\PageIndex{9A}\): Solving a Trigonometric Equation in Quadratic Form

Solve the equation exactly: \({\cos}^2 \theta+3 \cos \theta−1=0\), \(0≤\theta<2\pi\).

Solution

We begin by using substitution and replacing \(\cos \theta\) with \(x\). It is not necessary to use substitution, but it may make the problem easier to solve visually. Let \(\cos \theta=x\). We have

\(x^2+3x−1=0\)

The equation cannot be factored, so we will use the quadratic formula: \(x=\dfrac{−b\pm \sqrt{b^2−4ac}}{2a}\).

\[\begin{align*} x&= \dfrac{ -3\pm \sqrt{ {(-3)}^2-4 (1) (-1) } }{2}\\ &= \dfrac{-3\pm \sqrt{13}}{2}\end{align*}\]

Replace \(x\) with \(\cos \theta \) and solve.

\[\begin{align*} \cos \theta&= \dfrac{-3\pm \sqrt{13}}{2}\\ \theta&= {\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right) \end{align*}\]

Note that only the + sign is used. This is because we get an error when we solve \(\theta={\cos}^{−1}\left(\dfrac{−3−\sqrt{13}}{2}\right)\) on a calculator, since the domain of the inverse cosine function is \([ −1,1 ]\). However, there is a second solution:

\[\begin{align*} \theta&= {\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right)\\ &\approx 1.26 \end{align*}\]

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

\[\begin{align*} \theta&= 2\pi-{\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right)\\ &\approx 5.02 \end{align*}\]

Example \(\PageIndex{9B}\): Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: \(2 {\sin}^2 \theta−5 \sin \theta+3=0\), \(0≤\theta≤2\pi\).

Solution

Using grouping, this quadratic can be factored. Either make the real substitution, \(\sin \theta=u\),or imagine it, as we factor:

\[\begin{align*} 2 {\sin}^2 \theta-5 \sin \theta+3&= 0\\ (2 \sin \theta-3)(\sin \theta-1)&= 0 \qquad \text {Now set each factor equal to zero.}\\ 2 \sin \theta-3&= 0\\ 2 \sin \theta&= 3\\ \sin \theta&= \dfrac{3}{2}\\ \sin \theta-1&= 0\\ \sin \theta&= 1 \end{align*}\]

Next solve for \(\theta\): \(\sin \theta≠\dfrac{3}{2}\), as the range of the sine function is \([ −1,1 ]\). However, \(\sin \theta=1\), giving the solution \(\theta=\dfrac{\pi}{2}\).

Analysis

Make sure to check all solutions on the given domain as some factors have no solution.

Exercise \(\PageIndex{5}\)

Solve \({\sin}^2 \theta=2 \cos \theta+2\), \(0≤\theta≤2\pi\). [Hint: Make a substitution to express the equation only in terms of cosine.]

Answer

\(\cos \theta=−1\), \(\theta=\pi\)

Example \(\PageIndex{10A}\): Solving a Trigonometric Equation Using Algebra

Solve exactly: \(2 {\sin}^2 \theta+\sin \theta=0;\space 0≤\theta<2\pi\)

Solution

This problem should appear familiar as it is similar to a quadratic. Let \(\sin \theta=x\). The equation becomes \(2x^2+x=0\). We begin by factoring:

\[\begin{align*}
2x^2+x&= 0\\
x(2x+1)&= 0\qquad \text {Set each factor equal to zero.}\\
x&= 0\\
2x+1&= 0\\
x&= -\dfrac{1}{2} \end{align*}\]
Then, substitute back into the equation the original expression \(\sin \theta \) for \(x\). Thus,
\[\begin{align*} \sin \theta&= 0\\
\theta&= 0,\pi\\
\sin \theta&= -\dfrac{1}{2}\\
\theta&= \dfrac{7\pi}{6},\dfrac{11\pi}{6}
\end{align*}\]

The solutions within the domain \(0≤\theta<2\pi\) are \(\theta=0,\pi,\dfrac{7\pi}{6},\dfrac{11\pi}{6}\).

If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

\[\begin{align*} {\sin}^2 \theta+\sin \theta&= 0\\ \sin \theta(2\sin \theta+1)&= 0\\ \sin \theta&= 0\\ \theta&= 0,\pi\\ 2 \sin \theta+1&= 0\\ 2\sin \theta&= -1\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\dfrac{11\pi}{6} \end{align*}\]

Analysis

We can see the solutions on the graph in Figure \(\PageIndex{3}\). On the interval \(0≤\theta<2\pi\),the graph crosses the \(x\)-axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.

Figure \(\PageIndex{3}\)

We can verify the solutions on the unit circle in via the result in the section on Sum and Difference Identities as well.

Example \(\PageIndex{10B}\): Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in form exactly: \(2 {\sin}^2 \theta−3 \sin \theta+1=0\), \(0≤\theta<2\pi\).

Solution

We can factor using grouping. Solution values of \(\theta\) can be found on the unit circle.

\[\begin{align*} (2 \sin \theta-1)(\sin \theta-1)&= 0\\ 2 \sin \theta-1&= 0\\ \sin \theta&= \dfrac{1}{2}\\ \theta&= \dfrac{\pi}{6}, \dfrac{5\pi}{6}\\ \sin \theta&= 1\\ \theta&= \dfrac{\pi}{2} \end{align*}\]

Exercise \(\PageIndex{6}\)

Solve the quadratic equation \(2{\cos}^2 \theta+\cos \theta=0\).

Answer

\(\dfrac{\pi}{2}, \space \dfrac{2\pi}{3}, \space \dfrac{4\pi}{3}, \space \dfrac{3\pi}{2}\)