Question

practice assignment 8.3 inverse trigonometric funct
score: 6/11 answered: 6/11
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question 7
evaluate the expression \\( \sin ^ { - 1 } \left( \cos \left( \frac { 7 \pi } { 6 } \
ight) \
ight) \\).
give your answer as an exact value
question help: video 1 video 2
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Explanation:

Step1: Find the value of $\cos(\frac{7\pi}{6})$

Use the unit - circle or the formula $\cos(A + B)=\cos A\cos B-\sin A\sin B$. We know that $\frac{7\pi}{6}=\pi+\frac{\pi}{6}$.
By the formula $\cos(\pi + \alpha)=-\cos\alpha$, so $\cos(\frac{7\pi}{6})=\cos(\pi+\frac{\pi}{6})=-\cos(\frac{\pi}{6})=-\frac{\sqrt{3}}{2}$.

Step2: Evaluate $\sin^{-1}(-\frac{\sqrt{3}}{2})$

Let $y = \sin^{-1}(x)$, where $x=-\frac{\sqrt{3}}{2}$ and $y\in[-\frac{\pi}{2},\frac{\pi}{2}]$.
We know that $\sin(-\frac{\pi}{3})=-\sin(\frac{\pi}{3})=-\frac{\sqrt{3}}{2}$.

Answer:

$-\frac{\pi}{3}$