Example 2 from Algebra and Trigonometry, 8.3 Inverse Trigonometric Functions

Evaluating Inverse Trigonometric Functions for Special Input Values

Evaluate each of the following.

ⓐ ${sin}^{- 1} (\frac{1}{2})$
ⓑ ${sin}^{- 1} (- \frac{\sqrt{2}}{2})$
ⓒ $\cos^{- 1} (- \frac{\sqrt{3}}{2})$
ⓓ $\tan^{- 1} (1)$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

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Solution
1. ⓐ Evaluating $\sin^{- 1} (\frac{1}{2})$ is the same as determining the angle that would have a sine value of $\frac{1}{2} .$ In other words, what angle $x$ would satisfy $\sin (x) = \frac{1}{2} ?$ There are multiple values that would satisfy this relationship, such as $\frac{5 π}{4}$ and $\frac{5 π}{6},$ but we know we need the angle in the interval $[- \frac{π}{2}, \frac{π}{2}],$ so the answer will be $\sin^{- 1} (\frac{1}{2}) = \frac{π}{6} .$ Remember that the inverse is a function, so for each input, we will get exactly one output.
2. ⓑ To evaluate $\sin^{- 1} (- \frac{\sqrt{2}}{2}),$ we know that $\frac{5 π}{4}$ and $\frac{7 π}{4}$ both have a sine value of $- \frac{\sqrt{2}}{2},$ but neither is in the interval $[- \frac{π}{2}, \frac{π}{2}] .$ For that, we need the negative angle coterminal with $\cos^{- 1} (- \frac{\sqrt{3}}{2}) = \frac{5 π}{6} .$ ${sin}^{- 1} (- \frac{\sqrt{2}}{2}) = - \frac{π}{4} .$
3. ⓒTo evaluate $\cos^{- 1} (- \frac{\sqrt{3}}{2}),$ we are looking for an angle in the interval $[0, π]$ with a cosine value of $- \frac{\sqrt{3}}{2} .$ The angle that satisfies this is $\cos^{- 1} (- \frac{\sqrt{3}}{2}) = \frac{5 π}{6} .$
4. ⓓ Evaluating $\tan^{- 1} (1),$ we are looking for an angle in the interval $(- \frac{π}{2}, \frac{π}{2})$ with a tangent value of 1. The correct angle is $\tan^{- 1} (1) = \frac{π}{4} .$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Evaluate each of the following.

ⓐ ${sin}^{−1} (−1)$
ⓑ $\tan^{−1} (−1)$
ⓒ $\cos^{−1} (−1)$
ⓓ $\cos^{−1} (\frac{1}{2})$

How did it go?