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, $$ \begin{align} \cos^{-1}(2x^2-1)-2\sin^{-1}(x) &=\pi+2u-2u\\ &=\pi \end{align} $$

Answer 2

$\cos^{-1}x=y\implies\dfrac\pi2<y\le\pi$

$\cos^{-1}(\cos2y)=2y$ for $0\le y\le\dfrac\pi2$

and $=2\pi-2y$ for $y>\dfrac\pi2$

$\sin^{-1}x=\dfrac\pi2-y$

Answer 3

Show that the first derivative of the given function is zero on $[-1,0]$. So it is a constant function.

Answer 4

Let $y:=\arcsin x\in[-\frac{\pi}{2},\,0)$ so $x=\sin y$ and $\arccos(2x^2-1)=\arccos(-\cos 2y)=\pi+2y$, so the function is $\pi$.