Question

Prove that
sin(A ±B)=sin Acos B±cos AsinB​

Answer 1

Answer:

Step-by-step explanation:

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

cos(A+B)+isin(A+B)≡ei(A+B)≡eiA×eiB

≡[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)]

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

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