Question

The value of ${\cos }^2\left(\frac{\mathrm{\pi }}{6}+x\right)-{\sin }^2\left(\frac{\mathrm{\pi }}{6}-x\right)$ is
(a) $\frac{1}{2}\cos 2x$
(b) 0
(c) $-\frac{1}{2}\cos 2x$
(d) $\frac{1}{2}$

Answer 1

(a) $\frac{1}{2}\cos 2x$
$$\begin{gathered} {\cos }^2\left(\frac{\mathrm{\pi }}{6}+x\right)-{\sin }^2\left(\frac{\mathrm{\pi }}{6}-x\right) \\ =\cos \left(\frac{\mathrm{\pi }}{6}+x+\frac{\mathrm{\pi }}{6}-x\right)\cos \left(\frac{\mathrm{\pi }}{6}+x-\frac{\mathrm{\pi }}{6}+x\right)\left[\mathrm{Using}\cos (A+B)\cos (A-B)={\cos }^{2}A-{\sin }^{2}B\right] \\ =\cos \frac{2\mathrm{\pi }}{6}\cos 2x \\ =\frac{1}{2}\cos 2x\left[\mathrm{As}\cos \frac{\mathrm{\pi }}{3}=\frac{1}{2}\right] \end{gathered}$$