Evaluate $sin^{-1}(sin10)$

Solution :

We know that $sin^{-1}(sinx)$ = x, if $-\pi\over 2$ $\le$ x $\le$ $\pi\over 2$

Here, x = 10 radians which does not lie between -$\pi\over 2$ and $\pi\over 2$

But, $3\pi$ – x i.e. $3\pi$ – 10 lie between -$\pi\over 2$ and $\pi\over 2$

Also, sin($3\pi$ – 10) = sin 10

$\therefore$  $sin^{-1}(sin10)$ = $sin^{-1}(sin(3\pi – 10)$ = ($3\pi$ – 10)

---

Similar Questions

Solve the equation : 2$tan^{-1}({2x+1})$ = $cos^{-1}x$

Prove that : $sin^{-1}{12\over 13}$ + $cot^{-1}{4\over 3}$ + $tan^{-1}{63\over 16}$ = $\pi$

The value of $tan^{-1}(1)$ + $cos^{-1}({-1\over 2})$ + $sin^{-1}({-1\over 2})$ is equal to

Prove that : $cos^{-1}{12\over 13}$ + $sin^{-1}{3\over 5}$ = $sin^{-1}{56\over 65}$

Find the value of $sin^{-1}({-\sqrt{3}\over 2})$ + $cos^{-1}(cos({7\pi\over 6}))$.