Question

Let ${\cos }^{-1}(4x^3-3x)=a+b{\cos }^{-1}x$. If $x\in [-\frac{1}{2},\frac{1}{2}]$, then the principal value of ${\sin }^{-1}\left(\sin \frac{a}{b}\right)$ is

• A. $-\frac{\pi }{3}$
• B. $\frac{\pi }{3}$
• C. $-\frac{\pi }{6}$
• D. None of these
  

Answer 1

The correct option is D None of these

As $\cos 3x=4{\cos }^{3}x-3\cos x$

$\Rightarrow 3{\cos }^{-1}x={\cos }^{-1}(4x^3-3x)$

$\therefore a=0$ and $b=3$

or $a=2n\pi$ and $b=3.$

But it is given that $x\in [-\frac{1}{2},\frac{1}{2}]$ So, $a=0$ only

$\therefore {\sin }^{-1}\left(\sin \left(\frac{a}{b}\right)\right)={\sin }^{-1}\left(\sin \left(\frac{0}{3}\right)\right)=0$