Inverse Sine and cosine

Question

$\arcsin(\cos(x))=1/2$

Find $x$.

I got $-1/2$ or $2\pi-1/2$, but I don't know the correct answer. I tried graphing unit circle.

Answer 1

To begin, first remember that the sine and cosine functions are just phaseshifted copies of eachother, but otherwise behave the same. That is, $\sin(x)=\cos(x-\frac{\pi}{2})$ and that $\cos(x)=\sin(x+\frac{\pi}{2})$

$$\arcsin(\cos(x))=\frac{1}{2}$$

$$\arcsin(\sin(x+\frac{\pi}{2}))=\frac{1}{2}$$

$$x+\frac{\pi}{2} = \frac{1}{2}$$

$$x=\frac{1}{2}-\frac{\pi}{2}$$

So, one such answer is $\frac{1}{2}-\frac{\pi}{2}$. Remembering that $\cos(x)=\cos(2k\pi \pm x)$ for every integer $k$, the full list of possible answers then is $2k\pi\pm(\frac{1}{2}-\frac{\pi}{2})$.

Answer 2

on the unit circle, look at the points $1/2, \pi/2 - 1/2$ in the first quadrant. the terminal points have coordinates $(\cos 1/2, \sin 1/2), (\sin 1/2, \cos 1/2)$ mirror images on the mirror $y = x$. the $$\sin^{-1}(cos(1/2)) = \pi/2 - 1/2$$ because the terminal point corresponding to $\pi/2 - 1/2$ has the terminal point with the $y$-coordinate $\cos(1/2).$ other solutions are $1/2-\pi/2$ and you can add a multiple of $2\pi$ to either on of them to find all solutions.