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