Question

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

• A. $0$
• B. $\cos \,\,2\theta$
• C. $sin\,\,2\theta$
• D. $\cos \,\theta$

Answer 1

Answer: (A)

$0$

Answer 2

We know that, ${{\cos }^{2}}\,(A)-{{\sin }^{2}}(B)$
$=\cos (A+B)\,\cos \,(A-B)$
$\therefore$ ${{\cos }^{2}}\,\left( \frac{\pi }{4}+\theta \right)-{{\sin }^{2}}\left( \frac{\pi }{4}-\theta \right)$
$=\cos \,\left( \frac{\pi }{4}+\theta +\frac{\pi }{4}-\theta \right)\,\,\cos \,\left( \frac{\pi }{4}+\theta -\frac{\pi }{4}+\theta \right)$
$=\cos \,\left( \frac{\pi }{2} \right)\,\cos \,(2\,\theta )=0$