Question

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

Answer 1

sin(A+B)=sinAcosB+cosAsinB

Differentiating both sides with respect to x, we obtain
$\frac{d}{dx}$[sin(A+B)]=$\frac{d}{dx}$(sinAcosB)+$\frac{d}{dx}$(cosAsinB)
⇒cos(A+B).$\frac{d}{dx}$A+B)=cosB.$\frac{d}{dx}$(sinA)+sinA.$\frac{d}{dx}$cosB)+sinB.$\frac{d}{dx}$(cosA)+cosA.$\frac{d}{dx}$(sinB)
⇒cos(A+B).$\frac{d}{dx}$(A+B)=cosB.cosA$\frac{dA}{dx}$+sinA(-sinB)$\frac{dB}{dx}$+sinB(-sinA).$\frac{dA}{dx}$+cosAcosB$\frac{dB}{dx}$
⇒cos(A+B).[$\frac{dA}{dx}$+$\frac{dB}{dx}$]=(cosAcosb-sinAsinB).[$\frac{dA}{dx}$+$\frac{dB}{dx}$]
∴cos(A+B)=cosAcosB-sinAsinB