Question

cos(pi/2-x)cos(pi/2-y)-sin(pi/2-x)sin(pi/2-y)=-cos (x+y)

Answer 1

Answer:  The proof is done below.

Step-by-step explanation:  We are given to prove the following trigonometric equality :

$\cos\left(\dfrac{\pi}{2}-x\right)\cos\left(\dfrac{\pi}{2}-y\right)-\sin\left(\dfrac{\pi}{2}-x\right)\sin\left(\dfrac{\pi}{2}-y\right)=-\cos(x+y).$

We will be using the following formulas :

$(i)~\cos(A+B)=\cos A\cos B-\sin A\sin B,\\\\(ii)~\cos \left(\pi-A\right)=-\cos A.$

We have

$L.H.S.\\\\\\=\cos\left(\dfrac{\pi}{2}-x\right)\cos\left(\dfrac{\pi}{2}-y\right)-\sin\left(\dfrac{\pi}{2}-x\right)\sin\left(\dfrac{\pi}{2}-y\right)\\\\\\=\cos\left(\dfrac{\pi}{2}-x+\dfrac{\pi}{2}-y\right)~~~~~~~~~~~~~~~~~~~~~~~[\textup{Using formula (i)}]\\\\\\=\cos(\pi-(x+y))\\\\=-\cos(x+y)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Using formula (ii)}]\\\\=R.H.S.$

Hence proved.