If $x\in [-1,0)$, then what is the value of $\cos^{-1} (2x^2-1) – 2 \sin^{-1} x$

Question

If x belongs to $[-1,0)$ then what is $\cos^{-1} (2x^2-1) – 2 \sin^{-1} (x) $? This is a question in my book, but I have doubts about solving it. I tried putting $\sin a = x$ and got $\pi – 2a – 2a = \pi – 4a$. I'm supposed to get $\pi$ as the answer. How do I do this question?

Answer 1

Let $x=\sin(u)$, where $u\in\left[-\frac\pi2,0\right]$. Then $1-2x^2=\cos(2u)$. Then $2x^2-1=\cos(\pi+2u)$ and $\pi+2u\in[0,\pi]$. Thus, $$ \begin{align} \cos^{-1}(2x^2-1)-2\sin^{-1}(x) &=\pi+2u-2u\\ &=\pi \end{align} $$