If cos


If $\cos ^{2}\left(\frac{\pi}{6}+x\right)-\sin ^{2}\left(\frac{\pi}{6}-x\right)=k \cos 2 x$ then $k=$ __________________ 



$\cos ^{2}\left(\frac{\pi}{6}+x\right)-\sin ^{2}\left(\frac{\pi}{6}-x\right)=K \cos 2 x$

L.H.S. is $\left(\cos \left(\frac{\pi}{6}+x\right)\right)^{2}-\left(\sin \left(\frac{\pi}{6}-x\right)\right)^{2}$

$=\left(\cos \frac{\pi}{6} \cos x-\sin \frac{\pi}{6} \sin x\right)^{2}-\left(\sin \frac{\pi}{6} \cos x-\cos \frac{\pi}{6} \sin x\right)^{2}$

using identities : -

$\cos (a+b)=\cos a \cos b-\sin a \sin b$

and $\sin (a-b)=\sin a \cos b-\cos a \sin b$

$=\left(\frac{\sqrt{3}}{2} \cos x-\frac{1}{2} \sin x\right)^{2}-\left(\frac{1}{2} \cos x-\frac{\sqrt{3}}{2} \sin x\right)^{2}$

$=\frac{3}{4} \cos ^{2} x+\frac{1}{4} \sin ^{2} x-\frac{\sqrt{3}}{2} \sin x \cos x-\left[\frac{1}{4} \cos ^{2} x+\frac{3}{4} \sin ^{2} x-\frac{\sqrt{3}}{2} \sin x \cos x\right]$

$=\frac{3}{4} \cos ^{2} x-\frac{1}{4} \cos ^{2} x+\frac{1}{4} \sin ^{2} x-\frac{3}{4} \sin ^{2} x$

$=\frac{1}{2} \cos ^{2} x-\frac{1}{2} \sin ^{2} x$

$=\frac{1}{2}\left(\cos ^{2} x-\sin ^{2} x\right)$

$=\frac{1}{2} \cos 2 x=$ R.H.S. i. e. $K \cos 2 x$

$\Rightarrow K=\frac{1}{2}$

$\therefore$ value of $K$ is $\frac{1}{2}$.

Leave a comment