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