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7.2: Solving Trigonometric Equations with Identities

  • Page ID
    114041
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    Learning Objectives

    In this section, you will:

    • Verify the fundamental trigonometric identities.
    • Simplify trigonometric expressions using algebra and the identities.
    Photo of international passports.
    Figure 1 International passports and travel documents

    In espionage movies, we see international spies with multiple passports, each claiming a different identity. However, we know that each of those passports represents the same person. The trigonometric identities act in a similar manner to multiple passports—there are many ways to represent the same trigonometric expression. Just as a spy will choose an Italian passport when traveling to Italy, we choose the identity that applies to the given scenario when solving a trigonometric equation.

    In this section, we will begin an examination of the fundamental trigonometric identities, including how we can verify them and how we can use them to simplify trigonometric expressions.

    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 (see Table 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.

    Pythagorean Identities
    sin 2 θ+ cos 2 θ=1 sin 2 θ+ cos 2 θ=1 1+ cot 2 θ= csc 2 θ 1+ cot 2 θ= csc 2 θ 1+ tan 2 θ= sec 2 θ 1+ tan 2 θ= sec 2 θ
    Table 1

    The second and third identities can be obtained by manipulating the first. The identity 1+ cot 2 θ= csc 2 θ 1+ cot 2 θ= csc 2 θ is found by rewriting the left side of the equation in terms of sine and cosine.

    Prove: 1+ cot 2 θ= csc 2 θ 1+ cot 2 θ= csc 2 θ

    1+ cot 2 θ=( 1+ cos 2 θ sin 2 θ ) Rewrite the left side. =( sin 2 θ sin 2 θ )+( cos 2 θ sin 2 θ ) Write both terms with the common denominator. = sin 2 θ+ cos 2 θ sin 2 θ = 1 sin 2 θ = csc 2 θ 1+ cot 2 θ=( 1+ cos 2 θ sin 2 θ ) Rewrite the left side. =( sin 2 θ sin 2 θ )+( cos 2 θ sin 2 θ ) Write both terms with the common denominator. = sin 2 θ+ cos 2 θ sin 2 θ = 1 sin 2 θ = csc 2 θ

    Similarly, 1+ tan 2 θ= sec 2 θ 1+ tan 2 θ= sec 2 θ can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives

    1+ tan 2 θ=1+ ( sinθ cosθ ) 2 Rewrite left side. = ( cosθ cosθ ) 2 + ( sinθ cosθ ) 2 Write both terms with the common denominator. = cos 2 θ+ sin 2 θ cos 2 θ = 1 cos 2 θ = sec 2 θ 1+ tan 2 θ=1+ ( sinθ cosθ ) 2 Rewrite left side. = ( cosθ cosθ ) 2 + ( sinθ cosθ ) 2 Write both terms with the common denominator. = cos 2 θ+ sin 2 θ cos 2 θ = 1 cos 2 θ = sec 2 θ

    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 and determine whether the identity is odd or even. (See Table 2).

    Even-Odd Identities
    tan(θ)=tanθ cot(θ)=cotθ tan(θ)=tanθ cot(θ)=cotθ sin(θ)=sinθ csc(θ)=cscθ sin(θ)=sinθ csc(θ)=cscθ cos(θ)=cosθ sec(θ)=secθ cos(θ)=cosθ sec(θ)=secθ
    Table 2

    Recall that an odd function is one in which f(− x )= −f( x ) f(− x )= −f( x ) for all x x in the domain of f. f. The sine function is an odd function because sin( θ )=sinθ. sin( θ )=sinθ. The graph of an odd function is symmetric about the origin. For example, consider corresponding inputs of π 2 π 2 and π 2 . π 2 . The output of sin( π 2 ) sin( π 2 ) is opposite the output of sin( π 2 ). sin( π 2 ). Thus,

    sin( π 2 )=1 and sin( π 2 )=sin( π 2 ) =1 sin( π 2 )=1 and sin( π 2 )=sin( π 2 ) =1

    This is shown in Figure 2.

    Graph of y=sin(theta) from -2pi to 2pi, showing in particular that it is symmetric about the origin. Points given are (pi/2, 1) and (-pi/2, -1).
    Figure 2 Graph of y=sinθ y=sinθ

    Recall that an even function is one in which

    f( x )=f( x ) for all x in the domain of f 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(θ)=cosθ. cos(θ)=cosθ. For example, consider corresponding inputs π 4 π 4 and π 4 . π 4 . The output of cos( π 4 ) cos( π 4 ) is the same as the output of cos( π 4 ). cos( π 4 ). Thus,

    cos( π 4 )=cos( π 4 ) 0.707 cos( π 4 )=cos( π 4 ) 0.707

    See Figure 3.

    Graph of y=cos(theta) from -2pi to 2pi, showing in particular that it is symmetric about the y-axis. Points given are (-pi/4, .707) and (pi/4, .707).
    Figure 3 Graph of y=cosθ y=cosθ

    For all θ θ in the domain of the sine and cosine functions, respectively, we can state the following:

    • Since sin(−θ )=sinθ, sin(−θ )=sinθ, sine is an odd function.
    • Since, cos(− θ )=cosθ, cos(− θ )=cosθ, 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(− θ )=−tanθ. tan(− θ )=−tanθ. We can interpret the tangent of a negative angle as tan(− θ )= sin( θ ) cos(− θ ) = sinθ cosθ =tanθ. tan(− θ )= sin( θ ) cos(− θ ) = sinθ cosθ =tanθ. Tangent is therefore an odd function, which means that tan( θ )=tan( θ ) tan( θ )=tan( θ ) for all θ θ in the domain of the tangent function.

    The cotangent identity, cot( θ )=cotθ, cot( θ )=cotθ, also follows from the sine and cosine identities. We can interpret the cotangent of a negative angle as cot( θ )= cos( θ ) sin( θ ) = cosθ sinθ =cotθ. cot( θ )= cos( θ ) sin( θ ) = cosθ sinθ =cotθ. Cotangent is therefore an odd function, which means that cot( θ )=cot( θ ) cot( θ )=cot( θ ) for all θ θ 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( θ )= 1 sin( θ ) = 1 sinθ =cscθ. csc( θ )= 1 sin( θ ) = 1 sinθ =cscθ. 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( θ )= 1 cos( θ ) = 1 cosθ =secθ. sec( θ )= 1 cos( θ ) = 1 cosθ =secθ. 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. See Table 3.

    Reciprocal Identities
    sinθ= 1 cscθ sinθ= 1 cscθ cscθ= 1 sinθ cscθ= 1 sinθ
    cosθ= 1 secθ cosθ= 1 secθ secθ= 1 cosθ secθ= 1 cosθ
    tanθ= 1 cotθ tanθ= 1 cotθ cotθ= 1 tanθ cotθ= 1 tanθ
    Table 3

    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. See Table 4.

    Quotient Identities
    tanθ= sinθ cosθ tanθ= sinθ cosθ cotθ= cosθ sinθ cotθ= cosθ sinθ
    Table 4

    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 θ+ sin 2 θ=1 cos 2 θ+ sin 2 θ=1

    1+ cot 2 θ= csc 2 θ 1+ cot 2 θ= csc 2 θ

    1+ tan 2 θ= sec 2 θ 1+ tan 2 θ= sec 2 θ

    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( θ )=tanθ tan( θ )=tanθ

    cot( θ )=cotθ cot( θ )=cotθ

    sin( θ )=sinθ sin( θ )=sinθ

    csc( θ )=cscθ csc( θ )=cscθ

    cos( θ )=cosθ cos( θ )=cosθ

    sec( θ )=secθ sec( θ )=secθ

    The reciprocal identities define reciprocals of the trigonometric functions.

    sinθ= 1 cscθ sinθ= 1 cscθ

    cosθ= 1 secθ cosθ= 1 secθ

    tanθ= 1 cotθ tanθ= 1 cotθ

    cscθ= 1 sinθ cscθ= 1 sinθ

    secθ= 1 cosθ secθ= 1 cosθ

    cotθ= 1 tanθ cotθ= 1 tanθ

    The quotient identities define the relationship among the trigonometric functions.

    tanθ= sinθ cosθ tanθ= sinθ cosθ

    cotθ= cosθ sinθ cotθ= cosθ sinθ

    Example 1

    Graphing the Equations of an Identity

    Graph both sides of the identity cotθ= 1 tanθ . cotθ= 1 tanθ . In other words, on the graphing calculator, graph y=cotθ y=cotθ and y= 1 tanθ . y= 1 tanθ .

    Answer

    See Figure 4.

    Graph of y = cot(theta) and y=1/tan(theta) from -2pi to 2pi. They are the same!
    Figure 4

    Analysis

    We see only one graph because both expressions generate the same image. One is on top of the other. This is a good way to confirm an identity verified with analytical means. If both expressions give the same graph, then they are most likely identities.

    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 2

    Verifying a Trigonometric Identity

    Verify tanθcosθ=sinθ. tanθcosθ=sinθ.

    Answer

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

    tanθcosθ=( sinθ cosθ )cosθ =( sinθ cosθ ) cosθ =sinθ tanθcosθ=( sinθ cosθ )cosθ =( sinθ cosθ ) cosθ =sinθ

    Analysis

    This identity was fairly simple to verify, as it only required writing tanθ tanθ in terms of sinθ sinθ and cosθ. cosθ.

    Try It #1

    Verify the identity cscθcosθtanθ=1. cscθcosθtanθ=1.

    Example 3

    Verifying the Equivalency Using the Even-Odd Identities

    Verify the following equivalency using the even-odd identities:

    ( 1+sinx )[ 1+sin( x ) ]= cos 2 x ( 1+sinx )[ 1+sin( x ) ]= cos 2 x

    Answer

    Working on the left side of the equation, we have

    (1+sinx)[1+sin(−x)]=(1+sinx)(1sinx) Since sin(−x)=sinx =1 sin 2 x Difference of squares = cos 2 x cos 2 x=1 sin 2 x (1+sinx)[1+sin(−x)]=(1+sinx)(1sinx) Since sin(−x)=sinx =1 sin 2 x Difference of squares = cos 2 x cos 2 x=1 sin 2 x

    Example 4

    Verifying a Trigonometric Identity Involving sec2θ

    Verify the identity sec 2 θ1 sec 2 θ = sin 2 θ sec 2 θ1 sec 2 θ = sin 2 θ

    Answer

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

    sec 2 θ1 sec 2 θ = ( tan 2 θ+1)1 sec 2 θ sec 2 θ= tan 2 θ+1 = tan 2 θ sec 2 θ = tan 2 θ( 1 sec 2 θ ) = tan 2 θ( cos 2 θ) cos 2 θ= 1 sec 2 θ =( sin 2 θ cos 2 θ )( cos 2 θ) tan 2 θ= sin 2 θ cos 2 θ =( sin 2 θ cos 2 θ )( cos 2 θ ) = sin 2 θ sec 2 θ1 sec 2 θ = ( tan 2 θ+1)1 sec 2 θ sec 2 θ= tan 2 θ+1 = tan 2 θ sec 2 θ = tan 2 θ( 1 sec 2 θ ) = tan 2 θ( cos 2 θ) cos 2 θ= 1 sec 2 θ =( sin 2 θ cos 2 θ )( cos 2 θ) tan 2 θ= sin 2 θ cos 2 θ =( sin 2 θ cos 2 θ )( cos 2 θ ) = sin 2 θ

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

    sec 2 θ1 sec 2 θ = sec 2 θ sec 2 θ 1 sec 2 θ =1 cos 2 θ = sin 2 θ sec 2 θ1 sec 2 θ = sec 2 θ sec 2 θ 1 sec 2 θ =1 cos 2 θ = sin 2 θ

    Analysis

    In the first method, we used the identity sec 2 θ= tan 2 θ+1 sec 2 θ= tan 2 θ+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.

    Try It #2

    Show that cotθ cscθ =cosθ. cotθ cscθ =cosθ.

    Example 5

    Creating and Verifying an Identity

    Create an identity for the expression 2tanθsecθ 2tanθsecθ by rewriting strictly in terms of sine.

    Answer

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

    2tanθsecθ=2( sinθ cosθ )( 1 cosθ ) = 2sinθ cos 2 θ = 2sinθ 1 sin 2 θ Substitute 1 sin 2 θ for cos 2 θ 2tanθsecθ=2( sinθ cosθ )( 1 cosθ ) = 2sinθ cos 2 θ = 2sinθ 1 sin 2 θ Substitute 1 sin 2 θ for cos 2 θ

    Thus,

    2tanθsecθ= 2sinθ 1 sin 2 θ 2tanθsecθ= 2sinθ 1 sin 2 θ

    Example 6

    Verifying an Identity Using Algebra and Even/Odd Identities

    Verify the identity:

    sin 2 ( θ ) cos 2 ( θ ) sin( θ )cos( θ ) =cosθsinθ sin 2 ( θ ) cos 2 ( θ ) sin( θ )cos( θ ) =cosθsinθ

    Answer

    Let’s start with the left side and simplify:

    sin 2 ( θ ) cos 2 ( θ ) sin( θ )cos( θ ) = [ sin( θ ) ] 2 [ cos( θ ) ] 2 sin( θ )cos( θ ) = (− sinθ ) 2 ( cosθ ) 2 sinθcosθ sin(x)=sinxandcos(x)=cosx = ( sinθ ) 2 ( cosθ ) 2 sinθcosθ Difference of squares = ( sinθcosθ )( sinθ+cosθ ) ( sinθ+cosθ ) = ( sinθcosθ )( sinθ+cosθ ) ( sinθ+cosθ ) =cosθsinθ sin 2 ( θ ) cos 2 ( θ ) sin( θ )cos( θ ) = [ sin( θ ) ] 2 [ cos( θ ) ] 2 sin( θ )cos( θ ) = (− sinθ ) 2 ( cosθ ) 2 sinθcosθ sin(x)=sinxandcos(x)=cosx = ( sinθ ) 2 ( cosθ ) 2 sinθcosθ Difference of squares = ( sinθcosθ )( sinθ+cosθ ) ( sinθ+cosθ ) = ( sinθcosθ )( sinθ+cosθ ) ( sinθ+cosθ ) =cosθsinθ

    Try It #3

    Verify the identity sin 2 θ1 tanθsinθtanθ = sinθ+1 tanθ . sin 2 θ1 tanθsinθtanθ = sinθ+1 tanθ .

    Example 7

    Verifying an Identity Involving Cosines and Cotangents

    Verify the identity: ( 1 cos 2 x )( 1+ cot 2 x )=1. ( 1 cos 2 x )( 1+ cot 2 x )=1.

    Answer

    We will work on the left side of the equation.

    (1 cos 2 x)(1+ cot 2 x)=(1 cos 2 x)( 1+ cos 2 x sin 2 x ) =(1 cos 2 x)( sin 2 x sin 2 x + cos 2 x sin 2 x ) Find the common denominator. =(1 cos 2 x)( sin 2 x+ cos 2 x sin 2 x ) =( sin 2 x)( 1 sin 2 x ) =1 (1 cos 2 x)(1+ cot 2 x)=(1 cos 2 x)( 1+ cos 2 x sin 2 x ) =(1 cos 2 x)( sin 2 x sin 2 x + cos 2 x sin 2 x ) Find the common denominator. =(1 cos 2 x)( sin 2 x+ cos 2 x sin 2 x ) =( sin 2 x)( 1 sin 2 x ) =1

    Using Algebra to Simplify Trigonometric Expressions

    We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions before solving. Being familiar with the basic properties and formulas of algebra, such as the difference of squares formula, the perfect square formula, or substitution, will simplify the work involved with trigonometric expressions and equations.

    For example, the equation ( sinx+1 )( sinx1 )=0 ( sinx+1 )( sinx1 )=0 resembles the equation ( x+1 )( x1 )=0, ( x+1 )( x1 )=0, which uses the factored form of the difference of squares. Using algebra makes finding a solution straightforward and familiar. We can set each factor equal to zero and solve. This is one example of recognizing algebraic patterns in trigonometric expressions or equations.

    Another example is the difference of squares formula, a 2 b 2 =( ab )( a+b ), a 2 b 2 =( ab )( a+b ), which is widely used in many areas other than mathematics, such as engineering, architecture, and physics. We can also create our own identities by continually expanding an expression and making the appropriate substitutions. Using algebraic properties and formulas makes many trigonometric equations easier to understand and solve.

    Example 8

    Writing the Trigonometric Expression as an Algebraic Expression

    Write the following trigonometric expression as an algebraic expression: 2 cos 2 θ+cosθ1. 2 cos 2 θ+cosθ1.

    Answer

    Notice that the pattern displayed has the same form as a standard quadratic expression, a x 2 +bx+c. a x 2 +bx+c. Letting cosθ=x, cosθ=x, we can rewrite the expression as follows:

    2 x 2 +x1 2 x 2 +x1

    This expression can be factored as ( 2x1 )( x+1 ). ( 2x1 )( x+1 ). If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for x. x. At this point, we would replace x x with cosθ cosθ and solve for θ. θ.

    Example 9

    Rewriting a Trigonometric Expression Using the Difference of Squares

    Rewrite the trigonometric expression: 4 cos 2 θ1. 4 cos 2 θ1.

    Answer

    Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares. Thus,

    4 cos 2 θ1= (2cosθ) 2 1 =(2cosθ1)(2cosθ+1) 4 cos 2 θ1= (2cosθ) 2 1 =(2cosθ1)(2cosθ+1)

    Analysis

    If this expression were written in the form of an equation set equal to zero, we could solve each factor using the zero factor property. We could also use substitution like we did in the previous problem and let cosθ=x, cosθ=x, rewrite the expression as 4 x 2 1, 4 x 2 1, and factor ( 2x1 )( 2x+1 ). ( 2x1 )( 2x+1 ). Then replace x x with cosθ cosθ and solve for the angle.

    Try It #4

    Rewrite the trigonometric expression: 259 sin 2 θ. 259 sin 2 θ.

    Example 10

    Simplify by Rewriting and Using Substitution

    Simplify the expression by rewriting and using identities:

    csc 2 θ cot 2 θ csc 2 θ cot 2 θ

    Answer

    We can start with the Pythagorean identity.

    1+ cot 2 θ= csc 2 θ 1+ cot 2 θ= csc 2 θ

    Now we can simplify by substituting 1+ cot 2 θ 1+ cot 2 θ for csc 2 θ. csc 2 θ. We have

    csc 2 θ cot 2 θ=1+ cot 2 θ cot 2 θ =1 csc 2 θ cot 2 θ=1+ cot 2 θ cot 2 θ =1

    Try It #5

    Use algebraic techniques to verify the identity: cosθ 1+sinθ = 1sinθ cosθ . cosθ 1+sinθ = 1sinθ cosθ .

    (Hint: Multiply the numerator and denominator on the left side by 1sinθ.) 1sinθ.)

    Media

    Access these online resources for additional instruction and practice with the fundamental trigonometric identities.

    7.1 Section Exercises

    Verbal

    1.

    We know g(x)=cosx g(x)=cosx is an even function, and f(x)=sinx f(x)=sinx and h(x)=tanx h(x)=tanx are odd functions. What about G(x)= cos 2 x,F(x)= sin 2 x, G(x)= cos 2 x,F(x)= sin 2 x, and H(x)= tan 2 x? H(x)= tan 2 x? Are they even, odd, or neither? Why?

    2.

    Examine the graph of f(x)=secx f(x)=secx on the interval [π,π]. [π,π]. How can we tell whether the function is even or odd by only observing the graph of f(x)=secx? f(x)=secx?

    3.

    After examining the reciprocal identity for sect, sect, explain why the function is undefined at certain points.

    4.

    All of the Pythagorean Identities are related. Describe how to manipulate the equations to get from sin 2 t+ cos 2 t=1 sin 2 t+ cos 2 t=1 to the other forms.

    Algebraic

    For the following exercises, use the fundamental identities to fully simplify the expression.

    5.

    sinxcosxsecx sinxcosxsecx

    6.

    sin(x)cos(x)csc(x) sin(x)cos(x)csc(x)

    7.

    tanxsinx+secx cos 2 x tanxsinx+secx cos 2 x

    8.

    cscx+cosxcot(x) cscx+cosxcot(x)

    9.

    cott+tant sec(t) cott+tant sec(t)

    10.

    3 sin 3 tcsct+ cos 2 t+2cos(t)cost 3 sin 3 tcsct+ cos 2 t+2cos(t)cost

    11.

    tan(x)cot(x) tan(x)cot(x)

    12.

    sin(x)cosxsecxcscxtanx cotx sin(x)cosxsecxcscxtanx cotx

    13.

    1+ tan 2 θ csc 2 θ + sin 2 θ+ 1 sec 2 θ 1+ tan 2 θ csc 2 θ + sin 2 θ+ 1 sec 2 θ

    14.

    ( tanx csc 2 x + tanx sec 2 x )( 1+tanx 1+cotx ) 1 cos 2 x ( tanx csc 2 x + tanx sec 2 x )( 1+tanx 1+cotx ) 1 cos 2 x

    15.

    1 cos 2 x tan 2 x +2 sin 2 x 1 cos 2 x tan 2 x +2 sin 2 x

    For the following exercises, simplify the first trigonometric expression by writing the simplified form in terms of the second expression.

    16.

    tanx+cotx cscx ;cosx tanx+cotx cscx ;cosx

    17.

    secx+cscx 1+tanx ;sinx secx+cscx 1+tanx ;sinx

    18.

    cosx 1+sinx +tanx;cosx cosx 1+sinx +tanx;cosx

    19.

    1 sinxcosx cotx;cotx 1 sinxcosx cotx;cotx

    20.

    1 1cosx cosx 1+cosx ;cscx 1 1cosx cosx 1+cosx ;cscx

    21.

    ( secx+cscx )( sinx+cosx )2cotx;tanx ( secx+cscx )( sinx+cosx )2cotx;tanx

    22.

    1 cscxsinx ;secx and tanx 1 cscxsinx ;secx and tanx

    23.

    1sinx 1+sinx 1+sinx 1sinx ;secx and tanx 1sinx 1+sinx 1+sinx 1sinx ;secx and tanx

    24.

    tanx;secx tanx;secx

    25.

    secx;cotx secx;cotx

    26.

    secx;sinx secx;sinx

    27.

    cotx;sinx cotx;sinx

    28.

    cotx;cscx cotx;cscx

    For the following exercises, verify the identity.

    29.

    cosx cos 3 x=cosx sin 2 x cosx cos 3 x=cosx sin 2 x

    30.

    cosx( tanxsec( x ) )=sinx1 cosx( tanxsec( x ) )=sinx1

    31.

    1+ sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x =1+2 tan 2 x 1+ sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x =1+2 tan 2 x

    32.

    ( sinx+cosx ) 2 =1+2sinxcosx ( sinx+cosx ) 2 =1+2sinxcosx

    33.

    cos 2 x tan 2 x=2 sin 2 x sec 2 x cos 2 x tan 2 x=2 sin 2 x sec 2 x

    Extensions

    For the following exercises, prove or disprove the identity.

    34.

    1 1+cosx 1 1cos(x) =2cotxcscx 1 1+cosx 1 1cos(x) =2cotxcscx

    35.

    csc 2 x( 1+ sin 2 x )= cot 2 x csc 2 x( 1+ sin 2 x )= cot 2 x

    36.

    ( sec 2 (x) tan 2 x tanx )( 2+2tanx 2+2cotx )2 sin 2 x=cos2x ( sec 2 (x) tan 2 x tanx )( 2+2tanx 2+2cotx )2 sin 2 x=cos2x

    37.

    tanx secx sin( x )= cos 2 x tanx secx sin( x )= cos 2 x

    38.

    sec( x ) tanx+cotx =sin( x ) sec( x ) tanx+cotx =sin( x )

    39.

    1+sinx cosx = cosx 1+sin( x ) 1+sinx cosx = cosx 1+sin( x )

    For the following exercises, determine whether the identity is true or false. If false, find an appropriate equivalent expression.

    40.

    cos 2 θ sin 2 θ 1 tan 2 θ = sin 2 θ cos 2 θ sin 2 θ 1 tan 2 θ = sin 2 θ

    41.

    3 sin 2 θ+4 cos 2 θ=3+ cos 2 θ 3 sin 2 θ+4 cos 2 θ=3+ cos 2 θ

    42.

    secθ+tanθ cotθ+cosθ = sec 2 θ secθ+tanθ cotθ+cosθ = sec 2 θ


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