If cos²(π/6 + x) − sin²(π/6 − x) = k cos 2x, Find k
If \[ \cos^2\left(\frac{\pi}{6}+x\right) – \sin^2\left(\frac{\pi}{6}-x\right) = k\cos2x \] then \[ k= \] ………………………………………..
Solution
Using the identities:
\[ \cos^2\theta=\frac{1+\cos2\theta}{2} \]
and
\[ \sin^2\theta=\frac{1-\cos2\theta}{2} \]
Therefore,
\[ \cos^2\left(\frac{\pi}{6}+x\right) = \frac{ 1+\cos\left(\frac{\pi}{3}+2x\right) }{2} \]
Also,
\[ \sin^2\left(\frac{\pi}{6}-x\right) = \frac{ 1-\cos\left(\frac{\pi}{3}-2x\right) }{2} \]
Subtracting,
\[ = \frac{ \cos\left(\frac{\pi}{3}+2x\right) + \cos\left(\frac{\pi}{3}-2x\right) }{2} \]
Using the identity:
\[ \cos C+\cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \]
we get
\[ = \frac{ 2\cos\frac{2\pi/3}{2}\cos\frac{4x}{2} }{2} \]
\[ = \cos\frac{\pi}{3}\cos2x \]
Since
\[ \cos\frac{\pi}{3}=\frac12 \]
Therefore,
\[ \cos^2\left(\frac{\pi}{6}+x\right) – \sin^2\left(\frac{\pi}{6}-x\right) = \frac12\cos2x \]
Comparing with
\[ k\cos2x \]
we get
\[ \boxed{k=\frac12} \]