Question

Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.

Answer 1

Given, sin(A + B) = sinA.cosB + cosA.sinB
differentiate both sides with respect to x,
d[sin(A + B)]/dx = d[sinA.cosB]/dx + d[cosA.sinB]/dx

cos(A + B). d(A + B)/dx = sinA.d(cosB)/dx + cosB.d(sinA)/dx + cosA.d(sinB)/dx + sinB.d(cosA)/dx

cos(A + B) [ dA/dx + dB/dx ] = sinA.(-sinB).dB/dx + cosB.cosA.dA/dx + cosA.cosB.dB/dx + sinB.(-sinA).dA/dx

cos(A + B) [ dA/dx + dB/dx] = -sinA.sinB. dB/dx + cosA.cosB. dA/dx + cosA.cosB.dB/dx - sinA.sinB.dA/dx

cos(A + B).[dA/dx + dB/dx] = [dA/dx + dB/dx](cosA.cosB) - [dA/dx + dB/dx](sinA.sinB)

cos(A + B).[dA/dx + dB/dx] = (cosA.cosB - sinA.sinB)[dA/dx + dB/dx]

cos(A + B) = cosA.cosB - sinA.sinB