Discussion Forum

Let$f:(-1,1)\rightarrow R$ be such  that  $f(\cos 4 \theta)=\frac{2}{2 -\sec^{2} \theta}$  for 

$\theta \epsilon\left(0,\frac{\pi}{4}\right)\cup\left(\frac{\pi}{4},\frac{\pi}{2}\right)$ then, the value (s) of $f(\frac{1}{3})$ is/are 

Answer:

Explanation:

$f(\cos 4 \theta)=\frac{2}{2-\sec^{2} \theta}$ .......(i)

 at , $\cos 4 \theta=\frac{1}{3}$

 $\Rightarrow$   $2 \cos^{2} 2 \theta-1=\frac{1}{3}$

$\Rightarrow$   $\cos^{2} 2 \theta=\frac{2}{3}$

$\Rightarrow$    $\cos 2 \theta= \pm \sqrt{\frac{2}{3}}$ ........(ii)

 $\therefore$     $f(\cos 4 \theta)=\frac{2 \cos^{2} \theta}{2 \cos^{2} \theta-1}$

                           =$\frac{1+ \cos 2 \theta}{\cos 2 \theta}$

 $\Rightarrow$   $f\left(\frac{1}{3}\right)=1 \pm\sqrt{\frac{3}{2}}$ from eq (ii)