Question

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

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

Answer 1

The correct option is A

$\left(\frac{1}{2}\right)\cos 2\theta$

Explanation for the correct option:

Given, ${\cos }^2(\frac{\pi }{6}+\theta )–{\sin }^2(\frac{\pi }{6}–\theta )$

Use the identity $\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{A}\mathbf{}\mathbf{–}\mathbf{}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{B}\mathbf{)}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{A}\mathbf{}\mathbf{–}\mathbf{}\mathit{B}\mathbf{)}$,

$=\cos \left(\frac{\pi }{6}+\theta +\frac{\pi }{6}–\theta \right)\cos \left(\frac{\pi }{6}+\theta –\frac{\pi }{6}+\theta \right)$

$=\cos \left(\frac{2\pi }{6}\right)\cos 2\theta$

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

$=\left(\frac{1}{2}\right)\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{\theta }$

Hence, Option ‘A’ is Correct.