# 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)

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