Question

If A + B + C = $\frac{3\pi}{2}$, then prove that sin 2A + sin 2B - sin 2C = -4 sin A sin B sin C.

Answer 1

Answer:

Step-by-step explanation:

Formula used:

1.sinC - sinD= 2cos((C+D)/2)sin((C-D)/2)

2.sin2A= 2 sinA cosA

$sin2A+sin 2B\:-sin 2C\\=2sinA\:cosA+2\:cos(B+C).sin(B-C)\\=2sinA\:cosA+2\:cos(270-A).sin(B-C)\\=2sinA\:cosA-2\:sinA.sin(B-C)$

=2sinA[cosA-sin(B-C)]

=2sinA[cos(270-(B+C))-sin(B-C)]

=2sinA[-sin(B+C))-sin(B-C)]

=-2sinA[sin(B+C))+sin(B-C)]

= =-2sinA[2sinB.cosC]

=-4sinAsinB.cosC