Question

Prove: cos(π/2 + α-ß)= cosαsinß-sinαcosß

Answer 1

I'll use a for alpha and b for beta to save time. Start with this identity:\[\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\]Replace b with -b.\[\cos(a-b)=\cos(a)\cos(-b)-\sin(a)\sin(-b)\]Remember that cos(-b)=cos(b) because it's even and sin(-b)=-sin(b) because it's odd.\[\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)\]Now replace a with pi/2 + a.\[\cos(\frac{\pi}{2}+a-b)=\cos(\frac{\pi}{2}+a)\cos(b)+\sin(\frac{\pi}{2}+a)\sin(b)\]When you add pi/2 in the argument of sine or cosine, you shift it by pi/2 to the left. Therefore, sin(pi/2 + a) = cos(a) and cos(pi/2 + a) = -sin(a).\[\cos(\frac{\pi}{2}+a-b)=-\sin(a)\cos(b)+\cos(a)\sin(b)\]\[\cos(\frac{\pi}{2}+a-b)=\cos(a)\sin(b)-\sin(a)\cos(b)\]