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

We have sin (A + B) = sin A cos B + cos A sin B

Let A and B be functions of t.

Differentiating both sides with respect to t,

L.H.S. = `d/dx sin (A + B) = cos (A + B) ((dA)/dt + (dB)/dt)`

R.H.S. = `d/dt` (sin A cos B + cos A sin B)

`= cos A (dA)/dt cos B + sin A (- sin B) (dB)/dt + (- sin A) (dA)/dt sin B + cos A cos B (dB)/dt`

`= (cos A cos B - sin A sin B) (dA)/dt + (cos A cos B - sin A sin B) (dB)/dt`

`= (cos A cos B - sin A sin B)((dA)/dt + (dB)/dt)`

`= cos (A + B) ((dA)/dt + (dB)/dt)`

`= (cos A cos B - sin A sin B)((dA)/dt + (dB)/dt)`

Hence, cos (A + B) = cos A cos B – sin A sin B