Prove that,
sin(A+B) = SinACosB+CosASinB
Answers
Answered by
0
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)).
please Mark brainiest thank you
Similar questions