How to solve $\sin^{-1}(\sin16)$?

Question

On solving this $\sin^{-1}[\sin(16)]$ . I am finding both $(5π-16)$ and $(16-5π)$ to lie in the range of $[-π/2 , π/2]$ .

But my text book only mentions $(5π-16)$ as an answer.

So my question is, Is there any criteria or rule for getting the answer.

Please check me where I am getting it wrong.

Answer 1

You have

• $\sin(x)=-\sin(-x)$
• $\sin^{-1}(y) = -\sin^{-1}(-y)$ when $y \in [-1,1]$
• $\sin(x)=-\sin(x-\pi)=\sin(x-2\pi)=-\sin(x-3\pi)=\sin(x-4\pi)=-\sin(x-5\pi)$
• $\sin^{-1}(\sin(x))=x$ when $x \in [-\frac \pi2,\frac\pi2]$
• $16-5\pi\in [-\frac \pi2,\frac\pi2]$

Thus

• $\sin^{-1}(\sin(16))= -\sin^{-1}(\sin(16-5\pi)) = -(16-5\pi) = 5\pi-16$

The point to remember is that the sine function has period $2 \pi$, so something slightly more complicated happens if you only subtract an odd multiple of $\pi$.

Try to use a similar approach to $\cos^{-1}(\cos(16))$. You might get something like $\cos^{-1}(\cos(16))=\pi-\cos^{-1}(\cos(16-5\pi)) = \pi-(16-5\pi)= 6\pi-16$