Question

The value of ${\cos }^2\left(\frac{\pi }{4}+\theta \right)-{\sin }^2\left(\frac{\pi }{4}-\theta \right)$ is

• A. $0$
• B. $\cos \left(2\theta \right)$
• C. $\sin \left(2\theta \right)$
• D. $\cos \left(\theta \right)$

Answer 1

The correct option is A

$0$

Explanation for correct option

Given trigonometric function is ${\cos }^2\left(\frac{\pi }{4}+\theta \right)-{\sin }^2\left(\frac{\pi }{4}-\theta \right)$

$$\begin{gathered} {\cos }^2\left(\frac{\pi }{4}+\theta \right)-{\sin }^2\left(\frac{\pi }{4}-\theta \right) \\ =\cos \left(\frac{\pi }{4}+\theta +\frac{\pi }{4}-\theta \right)\cos \left(\frac{\pi }{4}+\theta -\frac{\pi }{4}+\theta \right)\left[\because {\cos }^2\left(A\right)-{\cos }^2\left(B\right)=\cos \left(A+B\right)\cos \left(A-B\right)\right] \\ =\cos \left(\frac{\pi }{2}\right)\cos \left(2\theta \right) \\ =0\left[\because \cos \left(\frac{\pi }{2}\right)=0\right] \end{gathered}$$

Hence, the correct option is $\mathrm{Option}\left(\mathrm{A}\right)$