TRIGONOMETRIC EQUATIONS

[Pages:29]CHAPTER

13

CHAPTER TABLE OF CONTENTS 13-1 First-Degree Trigonometric Equations 13-2 Using Factoring to Solve Trigonometric Equations 13-3 Using the Quadratic Formula to Solve Trigonometric Equations 13-4 Using Substitution to Solve Trigonometric Equations Involving More Than One Function 13-5 Using Substitution to Solve Trigonometric Equations Involving Different Angle Measures Chapter Summary Vocabulary Review Exercises Cumulative Review

518

TRIGONOMETRIC

EQUATIONS

The triangle is a rigid figure, that is, its shape cannot be changed without changing the lengths of its sides. This fact makes the triangle a basic shape in construction. The theorems of geometry give us relationships among the measures of the sides and angles of triangle. Precisely calibrated instruments enable surveyors to obtain needed measurements. The identities and formulas of trigonometry enable architects and builders to formulate plans needed to construct the roads, bridges, and buildings that are an essential part of modern life.

First-Degree Trigonometric Equations 519

13-1 FIRST-DEGREE TRIGONOMETRIC EQUATIONS

A trigonometric equation is an equation whose variable is expressed in terms of a trigonometric function value. To solve a trigonometric equation, we use the same procedures that we used to solve algebraic equations. For example, in the equation 4 sin u 1 5 5 7, sin u is multiplied by 4 and then 5 is added. Thus, to solve for sin u, first add the opposite of 5 and then divide by 4.

4 sin u 1 5 5 7 2 5 5 25

4 sin u 5 2

4

sin 4

u

5

2 4

sin u 5

1 2

We

know

that

sin

30?

5

12,

so

one

value

of

u

is

30?;

u 1

5

30.

We

also

know

that since sin u is positive in the second quadrant, there is a second-

quadrant angle, u2, whose sine is 12. Recall the relationship between an angle in

any quadrant to the acute angle called the reference angle. The following table

compares the degree measures of u from 290? to 360?, the radian measures of u from 2p2 to 2p, and the measure of its reference angle.

Fourth Quadrant

First Quadrant

Second Quadrant

Third Quadrant

Fourth Quadrant

Angle

Reference Angle

290? , u , 0? 2p2 , u , 0 2u 2u

0? , u , 90?

0

,

u

,

p 2

u

u

90? , u , 180?

p 2

,

u

,

p

180? 2 u p2u

180? , u , 270?

p

,

u

,

3p 2

u 2 180? u2p

270? , u , 360?

3p 2

,

u

,

2p

360? 2 u 2p 2 u

The

reference

angle

for

the

second-quadrant

angle

whose

sine

is

1 2

has

a

degree

measure

of

30?

and

u 2

5

180?

2

30?

or

150?.

Therefore,

sin

u 2

5

12.

For

0? u , 360?, the solution set of 4 sin u 1 5 5 7 is {30?, 150?}. In radian mea-

sure, the solution set is Up6 , 56pV. In the example given above, it was possible to give the exact value of u that

makes the equation true. Often it is necessary to use a calculator to find an

approximate value. Consider the solution of the following equation.

5 cos u 1 7 5 3

5 cos u 5 24

cos u 5 245

u 5 arccos A 254 B

520 Trigonometric Equations

When we use a calculator to find u, the calculator will return the value of the function y 5 arccos x whose domain is 0? x 180? in degree measure or 0 x p in radian measure.

In degree measure:

ENTER: 2nd

5)

COS1 (-) 4

ENTER

DISPLAY: c o s - 1 ( - 4 / 5 )

143.1301024

To the nearest degree, one value of u is

143?. In addition to this second-quadrant

angle, there is a third-quadrant angle such that cos u 5 245. To find this third-quadrant angle, find the reference angle for u.

Let R be the measure of the reference angle of the second-quadrant angle. That is, R is the acute angle such that cos u 5 2cos R.

y

143? 37?

x

R 5 180? 2 u 5 180? 2 143? 5 37?

The measure of the third-quadrant angle is:

u 5 R 1 180? u 5 37? 1 180? u 5 217?

For 0? u 360?, the solution set of 5 cos u 1 7 5 3 is {143?, 217?}. If the value of u can be any angle measure, then for all integral values of n, u 5 143 1 360n or u 5 217 1 360n.

Procedure

To solve a linear trigonometric equation:

1. Solve the equation for the function value of the variable.

2. Use a calculator or your knowledge of the exact function values to write one value of the variable to an acceptable degree of accuracy.

3. If the measure of the angle found in step 2 is not that of a quadrantal angle, find the measure of its reference angle.

4. Use the measure of the reference angle to find the degree measures of each solution in the interval 0? u , 360? or the radian measures of each solution in the interval 0 u , 2p.

5. Add 360n (n an integer) to the solutions in degrees found in steps 2 and 4 to write all possible solutions in degrees.Add 2pn (n an integer) to the solutions in radians found in steps 2 and 4 to write all possible solutions in radians.

First-Degree Trigonometric Equations 521

The following table will help you find the locations of the angles that satisfy trigonometric equations. The values in the table follow from the definitions of the trigonometric functions on the unit circle.

sin u 5 a cos u 5 a tan u 5 b

Sign of a and b (0 * |a| * 1, b 0)

1

2

Quadrants I and II Quadrants III and IV

Quadrants I and IV Quadrants II and III

Quadrants I and III Quadrants II and IV

EXAMPLE 1

Find the solution set of the equation 7 tan u 5 2 !3 1 tan u in the interval 0? u , 360?.

Solution

How to Proceed

(1) Solve the equation for tan u:

(2)

Since

tan

u

is

positive,

u 1

can

be

a

first-quadrant angle:

(3) Since u is a first-quadrant angle, R 5 u:

(4) Tangent is also positive in the third quadrant. Therefore, there is a third-quadrant angle such that

tan u 5 !33. In the third quadrant,

u 2

5

180?

1

R:

Answer The solution set is {30?, 210?}.

7 tan u 5 2 !3 1 tan u

6 tan u 5 2 !3

tan

u

5

!3 3

u 1

5

30?

R 5 30?

u 2

5

180?

1

R

u 2

5

180?

1

30?

5

210?

522 Trigonometric Equations

EXAMPLE 2

Find, to the nearest hundredth, all possible solutions of the following equation in radians:

3(sin A 1 2) 5 3 2 sin A

Solution

How to Proceed

(1) Solve the equation for sin A:

(2) Use a calculator to find one value of A (be sure that the calculator is in RADIAN mode):

3(sin A 1 2) 5 3 2 sin A 3 sin A 1 6 5 3 2 sin A 4 sin A 5 23 sin A 5 243

ENTER: 2nd SIN1 (-) 3 4 )

ENTER

DISPLAY: s i n - 1 ( - 3 / 4 )

-.848062079

(3) Find the reference angle:

(4) Sine is negative in quadrants III and IV. Use the reference angle to find a value of A in each of these quadrants:

(5) Write the solution set:

One value of A is 20.848.

R 5 2A 5 2(20.848) 5 0.848

In In

quadrant quadrant

III: IV:

A1 A2

5 5

p 1 0.848 3.99 2p 2 0.848 5.44

{3.99 1 2pn, 5.44 1 2pn} Answer

Note: When sin21 A 243 B was entered, the calculator returned the value in the

interval 2p2 # u # p2 , the range of the inverse of the sine function. This is the measure of a fourth-quadrant angle that is a solution of the equation. However, solutions are usually given as angle measures in radians between 0 and 2p plus multiples of 2p. Note that the value returned by the calculator is 5.43 1 2p (21) 20.85.

EXAMPLE 3

Find all possible solutions to the following equation in degrees:

1 2

(

sec

u

1

3)

5

sec

u

1

5 2

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