Question

prove that
sin (A+3B) + sin (3A+B) /
sin 2A + sin 2B = 2 cos (A+B)​

Answer 1

Answer:

3B+3A=6C

6C is the correct answer

Answer 2

Answer:

Step-by-step explanation:

$\frac{sin(A+3B)+sin(3A+B)}{sin2A+sin2B}$

=$\frac{2sin(2A+2B)cos(B-A)}{2sin(A+B)cos(A-B)}$

=$\frac{sin2(A+B)}{sin(A+B)}$

=$\frac{2sin(A+B)cos(A+B)}{sin(A+B)}$

=2 cos (A+B)