Q142 / 200 · TS EAMCET 2001 · Mathematics

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

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

Solution

$\begin{aligned} & \cos ^2\left(\frac{\pi}{6}+\theta\right)-\sin ^2\left(\frac{\pi}{6}-\theta\right) \\ & =\cos \left(\frac{\pi}{6}+\theta+\frac{\pi}{6}-\theta\right) \cos \left(\frac{\pi}{6}+\theta-\frac{\pi}{6}+\theta\right) \\ & =\cos \left(\frac{\pi}{3}\right) \cos 2 \theta=\frac{1}{2} \cos 2 \theta\end{aligned}$

Hence, the correct answer is option: 'A'.

Relevant Questions