6.8 Notes

In this lesson you will learn how to evaluate expressions containing trigonometric functions and inverse trigonometric relations.

1 3cos sin

2

6.8 Notes

From a prior lesson, you know there are an infinite number of angles for which

the sine of those angles is . In that

lesson, the instructions included an interval and most equations contained two solutions.

1 3cos sin

2

3

2

6.8 Notes

If the instructions for these expressions included an interval, then these expressions could possibly have two answers.But the instructions for these expressions are that you are to assume all angles lie in quadrant I; therefore, the expressions will have only one answer.

1 3cos sin

2

6.8 Notes

Example 1:

Simplify. Assume all angles lie in quadrant I

1

2

1 3cos sin

2

cos60

6.8 Notes

Example 2:

Simplify. Assume all angles lie in quadrant I

1

2

1 1cos cos

2

cos60

6.8 NotesExample 3:

Simplify. Assume all angles lie in quadrant I

1

sin arctan 3 cot 3arc

3sin 60 arctan

3

sin 60 30

sin 90

6.8 NotesExample 4:

Simplify. Assume all angles lie in quadrant I

2

1 1sec cos

2

sec60

1

cos 60112

6.8 Notes – practice problems:Simplify. Assume all angles lie in quadrant I

1 13 34. tan sin cos

2 2

3

3

2 22. tan arcsin cot arccos

2 2

13. tan sec 2

0

1

2

1 31. sin cos

2

3

6.8 Notes

In the prior example and practice problems, the argument of the inverse trigonometric relation expression has been an exact value of a trigonometric function of a special angle.

How will you simplify an expression where the argument is not an exact value of a trigonometric function of a special angle?

1 3cos sin

5

6.8 Notes

Similar to the way today’s “do now” was solved, you will draw and label a triangle, use the Pythagorean Theorem to find the third side measurement, and by knowing the ratios of sides, evaluate the expression. How will you know in which quadrant to draw the triangle?

The instructions will continue to be to assume that all angles lie in quadrant I.

6.8 NotesExample 5:Simplify. Assume all angles lie in quadrant I

Draw a triangle in quadrant I and label the opposite or “y” side 3 and the hypotenuse or “r” side 5. Use the Pythagorean Theorem to find the length of the adjacent or “x” side; it is 4.

1 3cos sin

5

6.8 Notes

Unlike prior example and practice problems, it is unnecessary to find thevalue of the angle whose sine is .

Knowing that cosine is the ratio of the adjacent side to the hypotenuse (x/r), theanswer is .

1 3cos sin

5

4

5

3

5

6.8 NotesExample 6:

Simplify. Assume all angles lie in quadrant I

1 8sin cot

15

15

17

6.8 NotesExample 7:

Simplify. Assume all angles lie in quadrant I

2csc cos

3arc

3

53 5

5

6.8 Notes – practice problems:Simplify. Assume all angles lie in quadrant I

722. sin arccos

97

13. tan sec 3

65

97

7

241 24

1. tan cos25

2 2

6.8 NotesUp to now, you have been evaluating expressions containing trigonometric functions and inverse trigonometric relations.

Now, you will learn how to evaluate expressions containing trigonometric functions and inverse trigonometric functions.

6.8 NotesRemember that inverse trigonometric functions are distinguished from trigonometric relations by the use of capital letters.

inverse trigonometric relations:

inverse trigonometric functions:

1cos

2arc

1cos

2Arc

1 1

2cos

1 1

2Cos

6.8 NotesAlso, remember that the inverse trigonometric functions were “created” by restricting the range of the inverse trigonometric relations.

Range of Arcsin:

Range of Arccos:

Range of Arctan: , 90 ,902 2

or

0, 0 ,180or

, 90 ,902 2

or

6.8 Notes

Because there are an infinite number of answers to an inverse trigonometric relation, the directions indicated that you are to assume that angles lie in quadrant I.

But there can only be one answer to an inverse trigonometric function. And since the ranges of the inverse trigonometric functions are not only between 0° and 90° (or 0 and π/2) it cannot be assumed that all angles lie in quadrant I.

6.8 NotesWhen evaluating expressions containing trigonometric functions and inverse trigonometric functions, only the one angle in range of the corresponding inverse trigonometric function is the answer.

6.8 Notes

Example 1:

The range of Arccos is .

120

1cos

2Arc

0 ,180

6.8 Notes

Example 2:

The range of Arcsin is .

3

3sin

2Arc

,2 2

6.8 Notes

Example 3:

The range of Tan-1 is .

6

1 3

3Tan

,2 2

6.8 Notes

Example 4:

0

tan cos 1Arc

tan180

6.8 Notes

Example 5:

30

cos cos330Arc

3cos

2Arc

6.8 Notes

Example 6:

Since is not the exact value of the

tangent of a special angle, a triangle is drawn to evaluate this expression.

Since the range of Arctan is ,

the triangle is drawn in quadrant IV where the Arctan of angles is negative.

1sin tan

2Arc

1

2

,2 2

6.8 Notes

Example 6:1

sin tan2

Arc

1

5

5

5

6.8 Notes

Example 7:

The hypotenuse must be the largest side; therefore, this triangle cannot exist.

It is not possible to evaluate this expression because is not in the domain of ; its domain is .

1 tan3

Sin

1 3Sin

3 1,1-1Sin

6.8 Notes – practice problems:

154. sin cos

17Arc

8

17

32. tan sin

2Arc

3. tan sin 2Arc

3

2

2 11. cos 1Tan

2 is not in the

domain of Arcsin