Finding the Exact Value of [latex] \sin ^{-1}(\sin x) \text { and } \cos ^{-1}(\cos x) [/latex] Find the exact value of a. [latex] \sin ^{-1}\left[\sin \left(-\frac{\pi}{8}\right)\right] [/latex] b. [latex] \cos ^{-1}\left(\cos \frac{5 \pi}{4}\right) [/latex]

a. Because [latex] -\frac{\pi}{2} \leq-\frac{\pi}{8} \leq \frac{\pi}{2}, \text { we have } \sin ^{-1}\left[\sin \left(-\frac{\pi}{8}\right)\right]=-\frac{\pi}{8}.[/latex] b. We cannot use the formula [latex] \cos ^{-1}(\cos x)=x \text { for } x=\frac{5 \pi}{4} \text { because } \frac{5 \pi}{4} [/latex] is not in the interval [latex] [0, \pi] . \text { However, } \cos \frac{5 \pi}{4}=\cos \left(2 \pi-\frac{5 \pi}{4}\right)=\cos \frac{3 \pi}{4} \text { and } \frac{3 \pi}{4} [/latex] is in the interval [latex][0, \pi] [/latex]. Therefore, [latex] \cos ^{-1}\left(\cos \frac{5 \pi}{4}\right)=\cos ^{-1}\left(\cos \frac{3 \pi}{4}\right)=\frac{3 \pi}{4} . [/latex]

Question 6.6.7: Finding the Exact Value of sin^-1 (sin x) and cos^-1 (cos x)...

Precalculus A Right Triangle Approach

Finding the Exact Value of $\sin ^{-1}(\sin x) \text { and } \cos ^{-1}(\cos x)$

Find the exact value of

a. $\sin ^{-1}\left[\sin \left(-\frac{\pi}{8}\right)\right]$ b. $\cos ^{-1}\left(\cos \frac{5 \pi}{4}\right)$

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a. Because $-\frac{\pi}{2} \leq-\frac{\pi}{8} \leq \frac{\pi}{2}, \text { we have } \sin ^{-1}\left[\sin \left(-\frac{\pi}{8}\right)\right]=-\frac{\pi}{8}.$

b. We cannot use the formula $\cos ^{-1}(\cos x)=x \text { for } x=\frac{5 \pi}{4} \text { because } \frac{5 \pi}{4}$ is not in the interval $[0, \pi] . \text { However, } \cos \frac{5 \pi}{4}=\cos \left(2 \pi-\frac{5 \pi}{4}\right)=\cos \frac{3 \pi}{4} \text { and } \frac{3 \pi}{4}$ is in the interval $[0, \pi]$ . Therefore,

$\cos ^{-1}\left(\cos \frac{5 \pi}{4}\right)=\cos ^{-1}\left(\cos \frac{3 \pi}{4}\right)=\frac{3 \pi}{4} .$