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

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.

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.

6.8 Notes Example 1: Simplify. Assume all angles lie in quadrant I

6.8 Notes Example 2: Simplify. Assume all angles lie in quadrant I

6.8 Notes Example 3: Simplify. Assume all angles lie in quadrant I

6.8 Notes Example 4: Simplify. Assume all angles lie in quadrant I

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

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?

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 Notes Example 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.

6.8 Notes Unlike prior example and practice problems, it is unnecessary to find the value of the angle whose sine is. Knowing that cosine is the ratio of the adjacent side to the hypotenuse (x/r), the answer is.

6.8 Notes Example 6: Simplify. Assume all angles lie in quadrant I

6.8 Notes Example 7: Simplify. Assume all angles lie in quadrant I

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

6.8 Notes Up 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 Notes Remember that inverse trigonometric functions are distinguished from trigonometric relations by the use of capital letters. inverse trigonometric relations: inverse trigonometric functions:

6.8 Notes Also, 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:

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 Notes When 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.

6.8 Notes Example 2: The range of Arcsin is.

6.8 Notes Example 3: The range of Tan -1 is.

6.8 Notes Example 4:

6.8 Notes Example 5:

6.8 Notes Example 6: Sinceis 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.

6.8 Notes Example 6:

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.

6.8 Notes – practice problems: