Let $x \in \mathbb R : \sin(\sin x )=1/2$ and $2<x<3$. What is the value of $\cos(-\sin x)$? Let $x \in \mathbb R : \sin(\sin x )=1/2$ and $2&lt;x&lt;3$.
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 :
$$\cos(-\sin(\sin^{-1}(\pi/6)))$$
Since $|\sin x|\le1$ for all $x$, and $1\lt\pi/2$, we have $\cos(\sin x)\gt0$, so that

$$\cos(-\sin x)=\cos(\sin x)=\sqrt{1-\sin^2(\sin x)}=\sqrt{1-\left(1\over2\right)^2}={\sqrt3\over2}$$
simply apply pythagorean identity and property of cosine function which is an even function.