Isn't it just ten?

$\sin ^{ -1 }{ \left( x \right) }$ lies in the range $-\frac { \pi }{ 2 } \quad to\quad \frac { \pi }{ 2 } \quad$ only. $\\ We\quad need\quad to\quad bring\quad \sin { \left( 10 \right) } \quad in\quad the\quad form\quad \sin { \left( k \right) } \\ where\quad k\quad lies\quad between\quad -\frac { \pi }{ 2 } \quad and\quad \frac { \pi }{ 2 } .\\ \\ \\ \sin { \left( -x \right) } =-\sin { \left( x \right) } \\ So\quad \sin { \left( 10 \right) } =-\sin { \left( -10 \right) } \qquad \qquad \qquad \qquad \qquad \qquad \left( 1 \right) \\ By\quad using\quad \sin { \left( a+b \right) } ,\quad we\quad can\quad show\quad that\\ \sin { \left( x+\left( 2n+1 \right) \pi \right) } =-\sin { \left( x \right) } \\ So\quad \sin { \left( -10+3\pi \right) } =-\sin { \left( -10 \right) } \qquad \qquad \qquad \qquad \quad \quad \left( 2 \right) \qquad \\ Using\quad \left( 1 \right) \quad and\quad \left( 2 \right) \\ \sin { \left( 10 \right) } =\sin { \left( -10+3\pi \right) } =\sin { \left( 3\pi -10 \right) } \\ \\ Here:\quad 3\pi -10\quad lies\quad in\quad the\quad range\quad of\quad \sin ^{ -1 }{ \left( x \right) } .\\ \\ So\quad \sin ^{ -1 }{ \left[ \sin { \left( 10 \right) } \right] } \\ =\boxed { 3\pi -10 }$

Recall that $\sin{x} = (-1)^k\sin({x}-k\pi)$ for all integers $k$ . Applying the odd function $\arcsin$ on both sides, we find that $\arcsin(\sin{x}) = (-1)^k\arcsin(\sin({x}-k\pi))$ . For $x=10$ we choose $k=3$ so that $x-k\pi$ falls in the interval $[-\pi/2,\pi/2]$ . Now $\arcsin(\sin{10}) = (-1)^3\arcsin(\sin({10}-3\pi))=-(10-3\pi)=3\pi-10$ .

In general, we have the formula $\arcsin(\sin{x}) = (-1)^k(x-k\pi)$ , where $k$ is the floor of $\frac{x+\pi/2}{\pi}$ .

use graph of sin^-1(sinx)

If k represents any integer then: $sinx = sin((2k+1)\pi - x)$ $\therefore\ sin^{-1} [sin(10)] = sin^{-1} [sin(3\pi - 10)] = 3\pi -10$

$\arcsin( \sin 10)$

= $\arcsin (\sin (10- 2\pi))$

= $\arcsin (\sin (\pi - (10 - 2\pi))$

= $arcsin (\sin(3\pi - 10))$

= $3\pi - 10$ since $\frac{-\pi}{2} \leq 3\pi - 10 \leq \frac{\pi}{2}$