Question

iii) Prove that SinA.Sin(60+A). Sin( 60- A) = 1/4 Sin3A​

Answer 1

Answer:

Step-by-step explanation:

L.H.S. = sin A sin (60 - A) sin (60 + A)

= sin A [ sin^2 60 - sin^2 A]      {sin (A + B) sin (A - B) = sin^2 A - sin^2 B }

= sin A [3/4 - sin^2 A]

= (3sin A)/4 - sin^3 A = 1/4 [3sin A - 4sin^3 A]

=1/4sin^3 A = R.H.S.