Question

${\cos }^2(\frac{\pi }{6}+\theta )-{\sin }^2(\frac{\pi }{6}-\theta )$ is equal to

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

Answer 1

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

As we know that

$\cos (A+B\cos (A-B)={\cos }^{2}A-{\sin }^{2}B$

${\cos }^2(\frac{\pi }{6}+\theta )-{\sin }^2(\frac{\pi }{6}-\theta )=\cos (\frac{\pi }{6}+\theta +\frac{\pi }{6}-\theta )\cos (\frac{\pi }{6}+\theta -\frac{\pi }{6}+\theta )$

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

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