Question

Geometrical proof of sin(A+B)

Answer 1

Here are two for the price of one (using Euler's formula):

cos(A+B)+isin(A+B)≡ei(A+B)≡eiA×eiBcos⁡(A+B)+isin⁡(A+B)≡ei(A+B)≡eiA×eiB

≡[cos(A)+isin(A)][cos(B)+isin(B)]≡[cos⁡(A)+isin⁡(A)][cos⁡(B)+isin⁡(B)]

≡[cos(A)cos(B)−sin(A)sin(B)]+i[sin(A)cos(B)+cos(A)sin(B)]≡[cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B)]+i[sin⁡(A)cos⁡(B)+cos⁡(A)sin⁡(B)]

Now equate imaginary parts to give the result for sin(A+B)sin⁡(A+B) (and, if you want, equate real parts to give the result for cos(A+B)cos⁡(A+B)).

Answer 2

Required Answer:-

In the geometrical proof of the addition formulae we are assuming that α, β and (α + β) are positive acute angles. But these formulae are true for any positive or negative values of α and β. Now we will prove that, sin (α + β) = sin α cos β + cos α sin β; where α and β are positive acute angles and α + β < 90°.

Hope it helps you ☺️