TITLE: Let $x \in \mathbb R : \sin(\sin x )=1/2$ and $2<x<3$. What is the value of $\cos(-\sin x)$?
QUESTION [1 upvotes]: Let $x \in \mathbb R : \sin(\sin x )=1/2$ and $2<x<3$.
What is the value of $\cos(-\sin x)$ ?

So far I have tried to solve:
$$\sin(\sin x)=1/2 \Leftrightarrow \sin x=\sin^{-1}(1/2) \Leftrightarrow \sin x=\pi/6 \Leftrightarrow x= \sin^{-1}(\pi/6) $$
Then I plugged it in :
$$\cos(-\sin(\sin^{-1}(\pi/6)))$$

REPLY [4 votes]: Since $|\sin x|\le1$ for all $x$, and $1\lt\pi/2$, we have $\cos(\sin x)\gt0$, so that
$$\cos(-\sin x)=\cos(\sin x)=\sqrt{1-\sin^2(\sin x)}=\sqrt{1-\left(1\over2\right)^2}={\sqrt3\over2}$$
{"set_name": "stack_exchange", "score": 1, "question_id": 2500408}