Question

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

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

Answer 1

Answer: (A)

${\cos }^2\left(\frac{\pi }{6}+\theta \right)-{\sin }^2\left(\frac{\pi }{6}-\theta \right)$
$=\cos \left(\frac{\pi }{6}+\theta +\frac{\pi }{6}-\theta \right)\cos \left(\frac{\pi }{6}+\theta -\frac{\pi }{6}+\theta \right)$
$=\cos \frac{2\pi }{6}\cos 2\theta =\frac{1}{2}\cos 2\theta$