Inverse trigonometric Problem

Question

For any $x \in [-1,0) \cup (0,1]$, how can I prove that: $$\sin^{-1}(2x\sqrt{1-x^2})=2\cos^{-1}x$$

Also, can someone explain to me how to understand the graphs of $sin$ and $cos$ functions?

Answer 1

Hint

Consider $x=\cos t$ and recall that $2\sin t\cos t=\sin 2t$.

Answer 2

Hint: If $\theta=2\cos^{-1}(x)$, then $x=\cos(\theta/2)$.

What is $\sqrt{1-x^2}$ ?

What is $2x\sqrt{1-x^2}$ ?

Answer 3

Sufficient care has to be taken while dealing with Inverse trigonometric functions

as the principal values of $\cos^{-1}x$ lies in $\left[0,\pi\right]$ whereas it lies in $\left[-\frac\pi2,\frac\pi2\right]$ for $\sin^{-1}y$

$\displaystyle\iff0\le2\cos^{-1}x\le2\pi$

Case $\#1:\displaystyle2\cos^{-1}x$ will be $\displaystyle\sin^{-1}2x\sqrt{1-x^2}$

iff $\displaystyle-\frac\pi2\le2\cos^{-1}x\le\frac\pi2\iff -\frac\pi4\le\cos^{-1}x\le\frac\pi4$

$\displaystyle\implies0\le\cos^{-1}x\le\frac\pi4\iff1\ge x\ge\frac1{\sqrt2}$

Case $\#2:$ If $\displaystyle\frac\pi4<\cos^{-1}x\le\frac\pi2\iff\frac1{\sqrt2}> x\ge0$

$\displaystyle\iff\frac\pi2<2\cos^{-1}x\le\pi,\sin^{-1}2x\sqrt{1-x^2}=\pi-2\cos^{-1}x$

Similarly, Case $\#3:$

$\displaystyle\iff\frac\pi2<\cos^{-1}x\le\frac{3\pi}4,\sin^{-1}2x\sqrt{1-x^2}=\pi-2\cos^{-1}x$

and Case $\#4:$

$\displaystyle\iff\frac{3\pi}4<\cos^{-1}x\le\pi,\sin^{-1}2x\sqrt{1-x^2}=2\pi-2\cos^{-1}x$

For example, if $\displaystyle x=-1,\cos^{-1}x=\cos^{-1}(-1)=\pi$

But, $\displaystyle\sin^{-1}\{2x\sqrt{1-x^2}\}=\sin^{-1}0=0$