Question

The value of $[\sin (x-\pi )+\cos (x-\frac{\pi }{2})].\cos (x-2\pi )$ _______

• A. $0$
• B. $1$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A $0$

we know that,

$\sin (-x)=-\sin x\cos (-x)=\cos x$

Now, $[\sin (x-\pi )+\cos (x-\frac{\pi }{2})].\cos (x-2\pi )=[\sin (-(\pi -x\left)\right)+\cos (-(\frac{\pi }{2}-x)\left)\right].\cos (-(2\pi -x\left)\right)=[-\sin (\pi -x)+\cos (\frac{\pi }{2}-x)].\cos (2\pi -x)=[-\sin x+\sin x].\cos x=0$

Therefore, Answer is $0$