Question

The value of ${\cos }^2\left(\frac{\pi }{6}+x\right)-{\sin }^2\left(\frac{\pi }{6}-x\right)$ is

• A. $\frac{1}{2}\cos 2x$
• B. $0$
• C. $-\frac{1}{2}\cos 2x$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{1}{2}\cos 2x$

$\cos \left(x+y\right)\cos \left(x-y\right)={\cos }^{2}x-{\sin }^{2}y$

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

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

$=\cos \frac{\pi }{3}\cos 2x$

$=\frac{1}{2}\cos 2x$