7.1|Solving Trigonometric Equations with Identities ⎝ π 2 ⎞ ⎠=1 and sin⎛ ⎝− π...

7.1 | Solving Trigonometric Equations with Identities

Learning Objectives

In this section, you will:

7.1.1 Verify the fundamental trigonometric identities.7.1.2 Simplify trigonometric expressions using algebra and the identities.

Figure 7.2 International passports and travel documents

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

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

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

To verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewritethe expression until it has been transformed into the same expression as the other side of the equation. Sometimes we haveto factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desiredresult. 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 7.1), which are equations involving trigonometric functionsbased on the properties of a right triangle. We have already seen and used the first of these identifies, but now we will alsouse additional identities.

894 Chapter 7 Trigonometric Identities and Equations

This content is available for free at http://legacy.cnx.org/content/col11667/1.4

Pythagorean Identities

sin2 θ + cos2 θ = 1 1 + cot2 θ = csc2 θ 1 + tan2 θ = sec2 θ

Table 7.1

The second and third identities can be obtained by manipulating the first. The identity 1 + cot2 θ = csc2 θ is found byrewriting the left side of the equation in terms of sine and cosine.

Prove: 1 + cot2 θ = csc2 θ

1 + cot2 θ = ⎛⎝1 + cos2 θsin2 θ⎞⎠ Rewrite the left side.

= ⎛⎝sin2 θsin2 θ⎞⎠+⎛⎝cos2 θ

sin2 θ⎞⎠ Write both terms with the common denominator.

= sin2 θ + cos2 θsin2 θ

= 1sin2 θ

= csc2 θ

Similarly, 1 + tan2 θ = sec2 θ can be obtained by rewriting the left side of this identity in terms of sine and cosine. Thisgives

1 + tan2 θ = 1 + ⎛⎝sin θcos θ⎞⎠

2Rewrite left side.

= ⎛⎝cos θcos θ⎞⎠

2+ ⎛⎝sin θ

cos θ⎞⎠

2Write both terms with the common denominator.

= cos2 θ + sin2 θcos2 θ

= 1cos2 θ

= sec2 θThe next set of fundamental identities is the set of even-odd identities. The even-odd identities relate the value of atrigonometric function at a given angle to the value of the function at the opposite angle and determine whether the identityis odd or even. (See Table 7.2).

Even-Odd Identities

tan( − θ) = − tan θcot( − θ) = − cot θ

sin( − θ) = − sin θcsc( − θ) = − csc θ

cos( − θ) = cos θsec( − θ) = sec θ

Table 7.2

Recall that an odd function is one in which f (−x)= − f (x) for all x in the domain of f . The sine function is an odd

function because sin(−θ) = − sin θ. The graph of an odd function is symmetric about the origin. For example, consider

corresponding inputs of π2 and − π2. The output of sin⎛⎝π2

⎞⎠ is opposite the output of sin⎛⎝−π

2⎞⎠. Thus,

Chapter 7 Trigonometric Identities and Equations 895

sin⎛⎝π2⎞⎠ = 1

and sin⎛⎝−π

2⎞⎠ = − sin⎛⎝π2

⎞⎠

= − 1

This is shown in Figure 7.3.

Figure 7.3 Graph of y = sin θ

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( − θ) = cos θ. For example, consider corresponding inputs π4 and − π4. The output of cos⎛⎝π4

⎞⎠ is the same as the

output of cos⎛⎝−π4⎞⎠. Thus,

cos⎛⎝−π4⎞⎠ = cos⎛⎝π4

⎞⎠

≈ 0.707

See Figure 7.4.

Figure 7.4 Graph of y = cos θ

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

• Since sin(−θ⎞⎠ = −sin θ, sine is an odd function.

• Since, cos(−θ⎞⎠ = cos θ, 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⎞⎠ = −tan θ. We can interpret the tangent of a negative angle as

tan(−θ⎞⎠ = sin(−θ)cos(−θ⎞⎠ = −sin θ

cos θ = − tan θ. Tangent is therefore an odd function, which means that tan(−θ) = − tan(θ)

for all θ in the domain of the tangent function.



The cotangent identity, cot(−θ) = − cot θ, also follows from the sine and cosine identities. We can interpret the

cotangent of a negative angle as cot(−θ) = cos(−θ)sin(−θ) = cos θ

−sin θ = − cot θ. Cotangent is therefore an odd function, which

means that cot(−θ) = − cot(θ) for all 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 beinterpreted as csc(−θ) = 1

sin(−θ) = 1−sin θ = − csc θ. 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) = 1cos θ = sec θ. 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, verifyingthe even-odd identities.

The next set of fundamental identities is the set of reciprocal identities, which, as their name implies, relate trigonometricfunctions that are reciprocals of each other. See Table 7.3.

Reciprocal Identities

sin θ = 1csc θ csc θ = 1

sin θ

cos θ = 1sec θ sec θ = 1

cos θ

tan θ = 1cot θ cot θ = 1

tan θ

Table 7.3

The final set of identities is the set of quotient identities, which define relationships among certain trigonometric functionsand can be very helpful in verifying other identities. See Table 7.4.

Quotient Identities

tan θ = sin θcos θ cot θ = cos θ

sin θ

Table 7.4

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.

(7.1)cos2 θ + sin2 θ = 1(7.2)1 + cot2 θ = csc2 θ(7.3)1 + tan2 θ = sec2 θ

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

(7.4)tan(−θ) = − tan θ


(7.5)cot(−θ) = − cot θ(7.6)sin(−θ) = − sin θ(7.7)csc(−θ) = − csc θ(7.8)cos(−θ) = cos θ(7.9)sec(−θ) = sec θ

The reciprocal identities define reciprocals of the trigonometric functions.

(7.10)sin θ = 1csc θ

(7.11)cos θ = 1sec θ

(7.12)tan θ = 1cot θ

(7.13)csc θ = 1sin θ

(7.14)sec θ = 1cos θ

(7.15)cot θ = 1tan θ

The quotient identities define the relationship among the trigonometric functions.

(7.16)tan θ = sin θcos θ

(7.17)cot θ = cos θsin θ

Example 7.1

Graphing the Equations of an Identity

Graph both sides of the identity cot θ = 1tan θ . In other words, on the graphing calculator, graph y = cot θ and

y = 1tan θ .

SolutionSee Figure 7.5.



Figure 7.5

AnalysisWe see only one graph because both expressions generate the same image. One is on top of the other. This is agood way to prove any identity. If both expressions give the same graph, then they must be identities.

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 tosimplify 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 makethe proper substitutions.

4. If these steps do not yield the desired result, try converting all terms to sines and cosines.

Example 7.2

Verifying a Trigonometric Identity

Verify tan θcos θ = sin θ.

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

tan θcos θ = ⎛⎝sin θcos θ⎞⎠cos θ

= ⎛⎝ sin θcos θ⎞⎠cos θ

= sin θ

AnalysisThis identity was fairly simple to verify, as it only required writing tan θ in terms of sin θ and cos θ.


7.1 Verify the identity csc θcos θtan θ = 1.

Example 7.3

Verifying the Equivalency Using the Even-Odd Identities

Verify the following equivalency using the even-odd identities:

(1 + sin x)⎡⎣1 + sin(−x)⎤⎦ = cos2 x

SolutionWorking on the left side of the equation, we have

(1 + sin x)[1 + sin(−x)] = (1 + sin x)(1 − sin x) Since sin(−x)= − sin x = 1 − sin2 x Difference of squares

= cos2 x cos2 x = 1 − sin2 x

Example 7.4

Verifying a Trigonometric Identity Involving sec2θ

Verify the identity sec2 θ − 1sec2 θ

= sin2 θ

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

sec2 θ − 1sec2 θ

= (tan2 θ + 1) − 1sec2 θ

sec2 θ = tan2 θ + 1

= tan2 θsec2 θ

= tan2 θ⎛⎝ 1sec2 θ⎞⎠

= tan2 θ(cos2 θ) cos2 θ = 1sec2 θ

= ⎛⎝sin2 θcos2 θ⎞⎠(cos2 θ) tan2 θ = sin2 θ

cos2 θ

=⎛⎝⎜ sin2 θ

cos2 θ

⎞⎠⎟(cos2 θ )

= sin2 θ

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



7.2

sec2 θ − 1sec2 θ

= sec2 θsec2 θ

− 1sec2 θ

= 1 − cos2 θ = sin2 θ

AnalysisIn the first method, we used the identity sec2 θ = tan2 θ + 1 and continued to simplify. In the second method, wesplit the fraction, putting both terms in the numerator over the common denominator. This problem illustrates thatthere 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.

Show that cot θcsc θ = cos θ.

Example 7.5

Creating and Verifying an Identity

Create an identity for the expression 2tan θsec θ by rewriting strictly in terms of sine.

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

2 tan θsec θ = 2⎛⎝sin θcos θ⎞⎠⎛⎝ 1cos θ⎞⎠

= 2 sin θcos2 θ

= 2 sin θ1 − sin2 θ

Substitute 1 − sin2 θ for cos2 θ

Thus,

2tan θsec θ = 2 sin θ1 − sin2 θ

Example 7.6

Verifying an Identity Using Algebra and Even/Odd Identities

Verify the identity:

sin2 (−θ) − cos2 (−θ)sin(−θ) − cos(−θ) = cos θ − sin θ

Solution


7.3

Let’s start with the left side and simplify:

sin2 (−θ) − cos2 (−θ)sin(−θ) − cos(−θ) =

⎡⎣sin(−θ)⎤⎦2 − ⎡

⎣cos(−θ)⎤⎦2sin(−θ) − cos(−θ)

= (−sin θ⎞⎠2 − (cos θ)2

−sin θ − cos θ sin(−x) = − sin x and cos(−x) = cos x

= (sin θ)2 − (cos θ)2

−sin θ − cos θ Difference of squares

= (sin θ − cos θ)(sin θ + cos θ)−(sin θ + cos θ)

= (sin θ − cos θ)⎛⎝sin θ + cos θ ⎞⎠−⎛⎝sin θ + cos θ ⎞⎠

= cos θ − sin θ

Verify the identity sin2 θ − 1tan θsin θ − tan θ = sin θ + 1

tan θ .

Example 7.7

Verifying an Identity Involving Cosines and Cotangents

Verify the identity: ⎛⎝1 − cos2 x⎞⎠⎛⎝1 + cot2 x⎞⎠ = 1.

SolutionWe will work on the left side of the equation.

(1 − cos2 x)(1 + cot2 x) = (1 − cos2 x)⎛⎝1 + cos2 xsin2 x⎞⎠

= (1 − cos2 x)⎛⎝sin2 xsin2 x

+ cos2 xsin2 x⎞⎠ Find the common denominator.

= (1 − cos2 x)⎛⎝sin2 x + cos2 xsin2 x

⎞⎠

= (sin2 x)⎛⎝ 1sin2 x⎞⎠

= 1

Using Algebra to Simplify Trigonometric ExpressionsWe have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifyingtrigonometric expressions before solving. Being familiar with the basic properties and formulas of algebra, such asthe difference of squares formula, the perfect square formula, or substitution, will simplify the work involved withtrigonometric expressions and equations.

For example, the equation (sin x + 1)(sin x − 1) = 0 resembles the equation (x + 1)(x − 1) = 0, which uses the factoredform of the difference of squares. Using algebra makes finding a solution straightforward and familiar. We can set eachfactor equal to zero and solve. This is one example of recognizing algebraic patterns in trigonometric expressions orequations.



7.4

Another example is the difference of squares formula, a2 − b2 = (a − b)(a + b), which is widely used in many areasother than mathematics, such as engineering, architecture, and physics. We can also create our own identities by continuallyexpanding an expression and making the appropriate substitutions. Using algebraic properties and formulas makes manytrigonometric equations easier to understand and solve.

Example 7.8

Writing the Trigonometric Expression as an Algebraic Expression

Write the following trigonometric expression as an algebraic expression: 2cos2 θ + cos θ − 1.

Solution

Notice that the pattern displayed has the same form as a standard quadratic expression, ax2 + bx + c. Letting cos θ = x, we can rewrite the expression as follows:

2x2 + x − 1

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

Example 7.9

Rewriting a Trigonometric Expression Using the Difference of Squares

Rewrite the trigonometric expression: 4 cos2 θ − 1.

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

4 cos2 θ − 1 = (2 cos θ)2 − 1 = (2 cos θ − 1)(2 cos θ + 1)

AnalysisIf this expression were written in the form of an equation set equal to zero, we could solve each factor usingthe zero factor property. We could also use substitution like we did in the previous problem and let cos θ = x,

rewrite the expression as 4x2 − 1, and factor (2x − 1)(2x + 1). Then replace x with cos θ and solve for theangle.

Rewrite the trigonometric expression: 25 − 9 sin2 θ.

Example 7.10


7.5

Simplify by Rewriting and Using Substitution

Simplify the expression by rewriting and using identities:

csc2 θ − cot2 θ

SolutionWe can start with the Pythagorean identity.

1 + cot2 θ = csc2 θ

Now we can simplify by substituting 1 + cot2 θ for csc2 θ. We have

csc2 θ − cot2 θ = 1 + cot2 θ − cot2 θ = 1

Use algebraic techniques to verify the identity: cos θ1 + sin θ = 1 − sin θ

cos θ .

(Hint: Multiply the numerator and denominator on the left side by 1 − sin θ.)

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

• Fundamental Trigonometric Identities (http://openstaxcollege.org/l/funtrigiden)

• Verifying Trigonometric Identities (http://openstaxcollege.org/l/verifytrigiden)



http://openstaxcollege.org/l/funtrigiden

http://openstaxcollege.org/l/verifytrigiden

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

7.1 EXERCISESVerbal

We know g(x) = cos x is an even function, and f (x) = sin x and h(x) = tan x are odd functions. What about

G(x) = cos2 x, F(x) = sin2 x, and H(x) = tan2 x? Are they even, odd, or neither? Why?

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

only observing the graph of f (x) = sec x?

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

All of the Pythagorean identities are related. Describe how to manipulate the equations to get from sin2 t + cos2 t = 1 tothe other forms.

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

sin x cos x sec x

sin(−x)cos(−x)csc(−x)

tan xsin x + sec xcos2 x

csc x + cos xcot(−x)

cot t + tan tsec( − t)

3 sin3 t csc t + cos2 t + 2 cos( − t)cos t

−tan(−x)cot(−x)

−sin(−x)cos x sec x csc x tan xcot x

1 + tan2 θcsc2 θ

+ sin2 θ + 1sec2 θ

⎛⎝ tan x

csc2 x+ tan x

sec2 x⎞⎠⎛⎝1 + tan x

1 + cot x⎞⎠− 1

cos2 x

1 − cos2 xtan2 x

+ 2 sin2 x

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

tan x + cot xcsc x ; cos x

sec x + csc x1 + tan x ; sin x


19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

cos x1 + sin x + tan x; cos x

1sin xcos x − cot x; cot x

11 − cos x − cos x

1 + cos x; csc x

(sec x + csc x)(sin x + cos x) − 2 − cot x; tan x

1csc x − sin x; sec x and tan x

1 − sin x1 + sin x − 1 + sin x

1 − sin x; sec x and tan x

tan x; sec x

sec x; cot x

sec x; sin x

cot x; sin x

cot x; csc x

For the following exercises, verify the identity.

cos x − cos3 x = cos x sin2 x

cos x⎛⎝tan x − sec(−x)⎞⎠ = sin x − 1

1 + sin2 xcos2 x

= 1cos2 x

+ sin2 xcos2 x

= 1 + 2 tan2 x

(sin x + cos x)2 = 1 + 2 sin xcos x

cos2 x − tan2 x = 2 − sin2 x − sec2 x

ExtensionsFor the following exercises, prove or disprove the identity.

11 + cos x − 1

1 − cos( − x) = − 2 cot x csc x

csc2 x⎛⎝1 + sin2 x⎞⎠ = cot2 x

⎛⎝sec2( − x) − tan2 x

tan x⎞⎠⎛⎝2 + 2 tan x

2 + 2 cot x⎞⎠− 2 sin2 x = cos 2x

tan xsec xsin(−x) = cos2 x

sec(−x)tan x + cot x = − sin(−x)



39.

40.

41.

42.

1 + sin xcos x = cos x

1 + sin(−x)

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

cos2 θ − sin2 θ1 − tan2 θ

= sin2 θ

3 sin2 θ + 4 cos2 θ = 3 + cos2 θ

sec θ + tan θcot θ + cos θ = sec2 θ


7.1|Solving Trigonometric Equations with Identities ⎝ π 2 ⎞ ⎠=1 and sin⎛ ⎝− π...

Documents

Transcript of 7.1|Solving Trigonometric Equations with Identities ⎝ π 2 ⎞ ⎠=1 and sin⎛ ⎝− π...