Question

Anybody know the right answer?

Answer 1

$\text{Use:}\\\\\sin(x-y)=\sin x\cos y-\sin y\cos x\\\\\cos(x-y)=\cos x\cos y+\sin x\sin y$

$\sin\left(\dfrac{\pi}{2}-x\right)=\sin\dfrac{\pi}{2}\cos x-\sin x\cos\dfrac{\pi}{2}=1\cos x-\sin x\cdot0=\cos x\\\\\cos\left(\dfrac{\pi}{2}-x\right)=\cos\dfrac{\pi}{2}\cos x+\sin\dfrac{\pi}{2}\sin x=0\cos x+1\sin x=\sin x$

$\sin\left(\dfrac{\pi}{2}-x\right)\cos x+\cos\left(\dfrac{\pi}{2}-x\right)\sin x\\\\=\cos x\cos x+\sin x\sin x=\cos^2x+\sin^2x=1$