7.1 Solving Trigonometric Equations with Identities - Precalculus 2e | OpenStax (2024)

Learning Objectives

In this section, you will:

  • Verify the fundamental trigonometric identities.
  • Simplify trigonometric expressions using algebra and the identities.
7.1 Solving Trigonometric Equations with Identities - Precalculus 2e | OpenStax (1)

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
sin2θ+cos2θ=1sin2θ+cos2θ=1 1+cot2θ=csc2θ1+cot2θ=csc2θ 1+tan2θ=sec2θ1+tan2θ=sec2θ

Table 1

The second and third identities can be obtained by manipulating the first. The identity 1+cot2θ=csc2θ1+cot2θ=csc2θ is found by rewriting the left side of the equation in terms of sine and cosine.

Prove: 1+cot2θ=csc2θ1+cot2θ=csc2θ

1+ cot 2 θ=( 1+ cos 2 θ sin 2 θ ) Rewritetheleftside. =( sin 2 θ sin 2 θ )+( cos 2 θ sin 2 θ ) Writebothtermswiththecommondenominator. = 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 ) Rewritetheleftside. =( sin 2 θ sin 2 θ )+( cos 2 θ sin 2 θ ) Writebothtermswiththecommondenominator. = sin 2 θ+ cos 2 θ sin 2 θ = 1 sin 2 θ = csc 2 θ

Similarly, 1+tan2θ=sec2θ1+tan2θ=sec2θ can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives

1+tan2θ=1+(sinθcosθ)2Rewriteleftside.=(cosθcosθ)2+(sinθcosθ)2Writebothtermswiththecommondenominator.=cos2θ+sin2θcos2θ=1cos2θ=sec2θ1+tan2θ=1+(sinθcosθ)2Rewriteleftside.=(cosθcosθ)2+(sinθcosθ)2Writebothtermswiththecommondenominator.=cos2θ+sin2θcos2θ=1cos2θ=sec2θ

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)=tanθcot(θ)=cotθtan(θ)=tanθcot(θ)=cotθ sin(θ)=sinθcsc(θ)=cscθsin(θ)=sinθcsc(θ)=cscθ cos(θ)=cosθsec(θ)=secθcos(θ)=cosθsec(θ)=secθ

Table 2

Recall that an odd function is one in which f(−x)= −f(x)f(−x)= −f(x) for all xx 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 π2π2 and π2.π2. The output of sin(π2)sin(π2) is opposite the output of sin(π2).sin(π2). 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.

7.1 Solving Trigonometric Equations with Identities - Precalculus 2e | OpenStax (2)

Figure 2 Graph of y=sinθy=sinθ

Recall that an even function is one in which

f(x)=f(x)forallxinthedomainofff(x)=f(x)forallxinthedomainoff

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 π4π4 and π4.π4. The output of cos(π4)cos(π4) is the same as the output of cos(π4).cos(π4). Thus,

cos(π4)=cos(π4)0.707cos(π4)=cos(π4)0.707

See Figure 3.

7.1 Solving Trigonometric Equations with Identities - Precalculus 2e | OpenStax (3)

Figure 3 Graph of y=cosθy=cosθ

For all θθ 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)=sinθcosθ=tanθ.tan(−θ)=sin(θ)cos(−θ)=sinθcosθ=tanθ. Tangent is therefore an odd function, which means that tan(θ)=tan(θ)tan(θ)=tan(θ) for all θθ 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)=cosθsinθ=cotθ.cot(θ)=cos(θ)sin(θ)=cosθsinθ=cotθ. Cotangent is therefore an odd function, which means that cot(θ)=cot(θ)cot(θ)=cot(θ) for all θθ 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)=1sin(θ)=1sinθ=cscθ.csc(θ)=1sin(θ)=1sinθ=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)=1cos(θ)=1cosθ=secθ.sec(θ)=1cos(θ)=1cosθ=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=1cscθsinθ=1cscθ cscθ=1sinθcscθ=1sinθ
cosθ=1secθcosθ=1secθ secθ=1cosθsecθ=1cosθ
tanθ=1cotθtanθ=1cotθ cotθ=1tanθcotθ=1tanθ

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=sinθcosθtanθ=sinθcosθ cotθ=cosθsinθcotθ=cosθsinθ

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.

cos2θ+sin2θ=1cos2θ+sin2θ=1

1+cot2θ=csc2θ1+cot2θ=csc2θ

1+tan2θ=sec2θ1+tan2θ=sec2θ

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)=tanθtan(θ)=tanθ

cot(θ)=cotθcot(θ)=cotθ

sin(θ)=sinθsin(θ)=sinθ

csc(θ)=cscθcsc(θ)=cscθ

cos(θ)=cosθcos(θ)=cosθ

sec(θ)=secθsec(θ)=secθ

The reciprocal identities define reciprocals of the trigonometric functions.

sinθ=1cscθsinθ=1cscθ

cosθ=1secθcosθ=1secθ

tanθ=1cotθtanθ=1cotθ

cscθ=1sinθcscθ=1sinθ

secθ=1cosθsecθ=1cosθ

cotθ=1tanθcotθ=1tanθ

The quotient identities define the relationship among the trigonometric functions.

tanθ=sinθcosθtanθ=sinθcosθ

cotθ=cosθsinθcotθ=cosθsinθ

Example 1

Graphing the Equations of an Identity

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

Solution

See Figure 4.

7.1 Solving Trigonometric Equations with Identities - Precalculus 2e | OpenStax (4)

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.

Solution

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 tanθtanθ in terms of sinθsinθ and cosθ.cosθ.

Example 3

Verifying the Equivalency Using the Even-Odd Identities

Verify the following equivalency using the even-odd identities:

(1+sinx)[ 1+sin(x) ]=cos2x(1+sinx)[ 1+sin(x) ]=cos2x

Solution

Working on the left side of the equation, we have

(1+sinx)[1+sin(−x)]=(1+sinx)(1sinx)Sincesin(−x)=sinx=1sin2xDifferenceofsquares=cos2xcos2x=1sin2x(1+sinx)[1+sin(−x)]=(1+sinx)(1sinx)Sincesin(−x)=sinx=1sin2xDifferenceofsquares=cos2xcos2x=1sin2x

Example 4

Verifying a Trigonometric Identity Involving sec2θ

Verify the identity sec2θ1sec2θ=sin2θsec2θ1sec2θ=sin2θ

Solution

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

sec2θ1sec2θ=(tan2θ+1)1sec2θsec2θ=tan2θ+1=tan2θsec2θ=tan2θ(1sec2θ)=tan2θ(cos2θ)cos2θ=1sec2θ=(sin2θcos2θ)(cos2θ)tan2θ=sin2θcos2θ=(sin2θcos2θ)(cos2θ)=sin2θsec2θ1sec2θ=(tan2θ+1)1sec2θsec2θ=tan2θ+1=tan2θsec2θ=tan2θ(1sec2θ)=tan2θ(cos2θ)cos2θ=1sec2θ=(sin2θcos2θ)(cos2θ)tan2θ=sin2θcos2θ=(sin2θcos2θ)(cos2θ)=sin2θ

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

sec2θ1sec2θ=sec2θsec2θ1sec2θ=1cos2θ=sin2θsec2θ1sec2θ=sec2θsec2θ1sec2θ=1cos2θ=sin2θ

Analysis

In the first method, we used the identity sec2θ=tan2θ+1sec2θ=tan2θ+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 cotθcscθ=cosθ.cotθcscθ=cosθ.

Example 5

Creating and Verifying an Identity

Create an identity for the expression 2tanθsecθ2tanθsecθ 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:

2tanθsecθ=2( sinθ cosθ )( 1 cosθ ) = 2sinθ cos 2 θ = 2sinθ 1 sin 2 θ Substitute1 sin 2 θfor cos 2 θ 2tanθsecθ=2( sinθ cosθ )( 1 cosθ ) = 2sinθ cos 2 θ = 2sinθ 1 sin 2 θ Substitute1 sin 2 θfor 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:

sin2(θ)cos2(θ)sin(θ)cos(θ)=cosθsinθsin2(θ)cos2(θ)sin(θ)cos(θ)=cosθsinθ

Solution

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 Differenceofsquares = ( 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 Differenceofsquares = ( sinθcosθ )( sinθ+cosθ ) ( sinθ+cosθ ) = ( sinθcosθ )( sinθ+cosθ ) ( sinθ+cosθ ) =cosθsinθ

Try It #3

Verify the identity sin2θ1tanθsinθtanθ=sinθ+1tanθ.sin2θ1tanθsinθtanθ=sinθ+1tanθ.

Example 7

Verifying an Identity Involving Cosines and Cotangents

Verify the identity: (1cos2x)(1+cot2x)=1.(1cos2x)(1+cot2x)=1.

Solution

We will work on the left side of the equation.

(1cos2x)(1+cot2x)=(1cos2x)(1+cos2xsin2x)=(1cos2x)(sin2xsin2x+cos2xsin2x)Findthecommondenominator.=(1cos2x)(sin2x+cos2xsin2x)=(sin2x)(1sin2x)=1(1cos2x)(1+cot2x)=(1cos2x)(1+cos2xsin2x)=(1cos2x)(sin2xsin2x+cos2xsin2x)Findthecommondenominator.=(1cos2x)(sin2x+cos2xsin2x)=(sin2x)(1sin2x)=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, a2b2=(ab)(a+b),a2b2=(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: 2cos2θ+cosθ1.2cos2θ+cosθ1.

Solution

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

2x2+x12x2+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 xx with cosθcosθ and solve for θ.θ.

Example 9

Rewriting a Trigonometric Expression Using the Difference of Squares

Rewrite the trigonometric expression: 4cos2θ1.4cos2θ1.

Solution

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,

4cos2θ1=(2cosθ)21=(2cosθ1)(2cosθ+1)4cos2θ1=(2cosθ)21=(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 4x21,4x21, and factor (2x1)(2x+1).(2x1)(2x+1). Then replace xx with cosθcosθ 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:

csc2θcot2θcsc2θcot2θ

Solution

We can start with the Pythagorean identity.

1+cot2θ=csc2θ1+cot2θ=csc2θ

Now we can simplify by substituting 1+cot2θ1+cot2θ for csc2θ.csc2θ. We have

csc2θcot2θ=1+cot2θcot2θ=1csc2θcot2θ=1+cot2θcot2θ=1

Try It #5

Use algebraic techniques to verify the identity: cosθ1+sinθ=1sinθcosθ.cosθ1+sinθ=1sinθcosθ.

(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.

  • Fundamental Trigonometric Identities
  • Verifying Trigonometric Identities

Verbal

1.

We know g(x)=cosxg(x)=cosx is an even function, and f(x)=sinxf(x)=sinx and h(x)=tanxh(x)=tanx are odd functions. What about G(x)=cos2x,F(x)=sin2x,G(x)=cos2x,F(x)=sin2x, and H(x)=tan2x?H(x)=tan2x? Are they even, odd, or neither? Why?

2.

Examine the graph of f(x)=secxf(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 sin2t+cos2t=1sin2t+cos2t=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+secxcos2xtanxsinx+secxcos2x

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+tan2θcsc2θ+sin2θ+1sec2θ1+tan2θcsc2θ+sin2θ+1sec2θ

14.

(tanxcsc2x+tanxsec2x)(1+tanx1+cotx)1cos2x(tanxcsc2x+tanxsec2x)(1+tanx1+cotx)1cos2x

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+cotxcscx;cosxtanx+cotxcscx;cosx

17.

secx+cscx1+tanx;sinxsecx+cscx1+tanx;sinx

18.

cosx1+sinx+tanx;cosxcosx1+sinx+tanx;cosx

19.

1sinxcosxcotx;cotx1sinxcosxcotx;cotx

20.

11cosxcosx1+cosx;cscx11cosxcosx1+cosx;cscx

21.

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

22.

1cscxsinx;secxandtanx1cscxsinx;secxandtanx

23.

1sinx1+sinx1+sinx1sinx;secxandtanx1sinx1+sinx1+sinx1sinx;secxandtanx

24.

tanx;secxtanx;secx

25.

secx;cotxsecx;cotx

26.

secx;sinxsecx;sinx

27.

cotx;sinxcotx;sinx

28.

cotx;cscxcotx;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))=sinx1cosx(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.

cos2xtan2x=2sin2xsec2xcos2xtan2x=2sin2xsec2x

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.

csc2x(1+sin2x)=cot2xcsc2x(1+sin2x)=cot2x

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.

tanxsecxsin(x)=cos2xtanxsecxsin(x)=cos2x

38.

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

39.

1+sinxcosx=cosx1+sin(x)1+sinxcosx=cosx1+sin(x)

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

40.

cos2θsin2θ1tan2θ=sin2θcos2θsin2θ1tan2θ=sin2θ

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+tanθcotθ+cosθ=sec2θsecθ+tanθcotθ+cosθ=sec2θ

7.1 Solving Trigonometric Equations with Identities - Precalculus 2e | OpenStax (2024)

FAQs

What is the trick to solve trigonometric identities? ›

How To: Given a trigonometric identity, verify that it is true.
  1. Work on one side of the equation. ...
  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.

How do you verify that the given trigonometric equation is an identity? ›

Graphing both sides of an identity will verify it. Simplifying one side of the equation to equal the other side is another method for verifying an identity. The approach to verifying an identity depends on the nature of the identity. It is often useful to begin on the more complex side of the equation.

How do you verify trigonometric identities expressions? ›

Change everything into terms of sine and cosine. Use the identities when you can. Start with simplifying the left-hand side of the equation, then, once you get stuck, simplify the right-hand side. As long as the two sides end up with the same final expression, the identity is true.

How to do trig identities quickly? ›

5 strategies you can use to solve TRIG IDENTITIES
  1. Multiply the denominator by a CONJUGATE.
  2. Get a COMMON DENOMINATOR.
  3. SPLIT UP A FRACTION into two separate fractions.
  4. Rewrite everything in terms of SINE AND COSINE.
Feb 23, 2017

What are the 7 trigonometric identities? ›

The following identities for the trigonometric ratio explain their periodicity.
  • sin (n × 360° + θ) = sin θ
  • sin (2nπ + θ) = sin θ
  • cos (n × 360° + θ) = cos θ
  • cos (2nπ + θ) = cos θ
  • tan (n × 180° + θ) = tan θ
  • tan (nπ + θ) = tan θ
  • cosec (n × 360° + θ) = cosec θ
  • cosec (2nπ + θ) = cosec θ
7 days ago

What is the trigonometric identities technique? ›

The general method of proving trigonometric identities is to work on each side of the equation separately, and simplify or manipulate each side until you reach the same expression on both sides. We're done once we've reached the same expression on both sides of the equation, specifically \(tanx\).

What is the T formula in trigonometry? ›

Introducing the parameter t=tanθ2 turns out to be a very useful tool in solving certain types of trigonometric equations and also in finding certain integrals involving trigonometric functions. The basic idea is to relate sinθ, cosθ and even tanθ to the tangent of half the angle.

How to solve trigonometric equations step by step? ›

Step 1: Move all terms to one side of the equation. Step 2: Rewrite using trigonometric identities until all trigonometric functions are the same type of function and have the same input. Step 3: Factor and set each factor equal to zero. Solve.

What is an identity in precalc? ›

A trigonometric equation is any equation that includes a trigonometric function. There are two basic types of trigonometric equations: identities and conditional equations. Identities are equations that hold for any angle. Conditional equations are equations that are solved only by certain angles.

How to simplify trigonometric identity? ›

When simplifying trigonometric expressions, one approach is to change everything into sine or cosine. First, we can change secant to cosine using the Reciprocal Identity. Now, combine the denominator into one fraction by multiplying 1 by ⁡ ⁡ . Change this problem into a division problem and simplify.

How do you identify a trigonometric equation? ›

Thus, to check to whether an equation is an identity, we can compare graphs of (left side of the equation) and (right side of the equation). If the two graphs are identical, the equation is an identity. If the two graphs are not the same, the equation is not an identity.

How do you evaluate trig identities? ›

To evaluate trigonometric expressions using identities, replace a complicated expression with a simpler one. If there are more than one trigonometric functions in the expression, try to replace them with a single function. Repeat this until the functions are simple to evaluate.

What is the trick for trig values? ›

The Trigonometry Hand Trick

The finger (or thumb) at the top is 0°, the next one down is 30°, then 45° and so on. To find the trig value of a particular angle, you start by putting down the finger for that angle. So, for 30° that would be your index finger.

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