Question

sin2A+sin2B+2sinAsinB cos(A+B) =sin2(A+B). prove that​

Answer 1

Answer:

Sin2A+Sin2B+Sin2(A−B)

Applying some trigonometric identity in above expression;

2 \times [ sin (2A+2B)/2 \times cos (2A-2B)/2] + sin 2(A-B)2×[sin(2A+2B)/2×cos(2A−2B)/2]+sin2(A−B)

2 \times [ sin (A+B) \times cos (A-B)] + sin 2(A-B)2×[sin(A+B)×cos(A−B)]+sin2(A−B)

2 \times sin (A+B) \times cos (A-B) + 2 \times sin (A-B)cos (A-B)2×sin(A+B)×cos(A−B)+2×sin(A−B)cos(A−B)

2 \times cos (A-B) [ sin (A+B) + sin (A-B)]2×cos(A−B)[sin(A+B)+sin(A−B)]

2 \times cos (A-B) \times  2 sin (A+B+A-B)/2 \times cos (A+B-A+B)/22×cos(A−B)× 2sin(A+B+A−B)/2×cos(A+B−A+B)/2

=  4 \times cos (A-B) \times sin A \times cos B4×cos(A−B)×sinA×cosB = R.H.S

∴ It is being proved that

=  sin 2A+sin 2B+sin 2(A-B)=4 sin A.cos B. cos(A-B)