Question

Find linear expression of $\sin^{-1}(\sin x)$ for $x$ belongs to $[2027\pi/2,2029\pi/2]$

Answer 1

$x-1014\pi$

Answer 2

For $\sin^{-1}(\sin x)$, use the formula:

$\sin^{-1}(\sin x)= (-1)^n (x - n\pi)$ for $x\in [n\pi-\frac{\pi}{2},\, n\pi+\frac{\pi}{2}]$.

For $x \in \left[\frac{2027\pi}{2}, \frac{2029\pi}{2}\right]$:

Determine $n$ such that:

$n\pi - \frac{\pi}{2} = \frac{2027\pi}{2}$ and $n\pi + \frac{\pi}{2} = \frac{2029\pi}{2}$.

Solving,

$n\pi - \frac{\pi}{2} = \frac{2027\pi}{2} \implies 2n\pi - \pi = 2027\pi \implies 2n\pi = 2028\pi \implies n = 1014$.

Since $n = 1014$ (an even number), $(-1)^{1014} = 1$. Therefore,

$\sin^{-1}(\sin x) = x - 1014\pi$.