Chapter 3 Inverse Trigonometric Functions

Problems Involving Inverse Trigonometric Functions

It is tempting to believe, for example, that sinsin or tantan . The two functions are, after all inverses. However, this is not always the case because the inverse function value desired is typically its principal value, which the student will recall is defined only in certain quadrants (see the table at right).

Let’s look at a couple of problems to see how they are solved.

Example 3.1: Calculate the principal value of tan tan . Begin by noticing that tan and tan are inverse functions, so the solution to this problem is related to the angle given: . This angle is in Q2, but the inverse tangent function is defined only in Q1 and Q4, on the interval , . We seek the angle in Q1 or Q4 that has the same tangent value as . Since the tangent function has period , we can calculate:

tan tan (in Q4) as our solution.

Example 3.2: Calculate the principal value of sin cos .

We are looking for the angle whose sine value is cos in the interval , .

Method 1 sin cos sin √ since sine values are negative in Q4. :

Method 2: Recall: sin θ cosθ. Then, cos sin sin . Then, sin cos sin sin b e c a u s e cos sin sin

sin sin because ≡ and is in the interval , . because inverse functions work nicely in quadrants in which the principal values of the inverse functions are defined.

Version 2.0 Page 35 of 109 January 1, 2016 Chapter 3 Inverse Trigonometric Functions

Problems Involving Inverse Trigonometric Functions

When the inverse trigonometric function is the inner function in a composition of functions, it will usually be necessary to draw a triangle to solve the problem. In these cases, draw the triangle defined by the inner (inverse trig) function. Then derive the value of the outer (trig) function.

Example 3.3: Calculate the value of cot sin √ . √ Recall that the argument of the sin function, . Draw the triangle based on this. Next, calculate the value of the triangle’s horizontal leg:

61 5√61 6√61. Based on the diagram, then,

5√61 6√61 6 cot sin 61 5√61 5

Example 3.4: Calculate the value of tan cos √ . √ Recall that the argument of the cos function, . Draw the triangle based on this. Next, calculate the value of the triangle’s vertical leg:

2 √2 √2.

Based on the diagram, then,

√2 √2 tan cos 1 2 √2

√ Example 3.5: Calculate an algebraic expression for sin sec . √ Recall that the argument of the sec function, . Draw the triangle based on this. Next, calculate the value of the triangle’s vertical leg:

√ 9 3

Based on the diagram, then,

√ 9 3 √ sin sec √ 9

Version 2.0 Page 36 of 109 January 1, 2016