Answer with explanation:
We are asked to prove the identity:
\(\cos (x+\dfrac{\pi}{2})=-\sin x\)
We know that:
\(\cos (A+B)=\cos A\cos B-\sin A\sin B\)
Here we have:
\(A=x\ and\ B=\dfrac{\pi}{2}\)
Hence,
\(\cos (x+\dfrac{\pi}{2})=\cos x\cos (\dfrac{\pi}{2})-\sin x\sin (\dfrac{\pi}{2})\)
We know that:
\(\cos \dfrac{\pi}{2}=0\\\\and\\\\\sin \dfrac{\pi}{2}=1\)
Hence, we have:
\(\cos (x+\dfrac{\pi}{2})=0-\sin x\\\\i.e.\\\\\cos (x+\dfrac{\pi}{2})=-\sin x\)
