Question

4 sin A sin (60° + A) sin (60° - A) = sin 3A.

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Answer 1

Proof: sin 3A

= sin (2A + A)

= sin 2A cos A + cos 2A sin A

= 2 sin A cos A ∙ cos A + (1 - 2 sin^2 A) sin A

= 2 sin A (1 - sin^2 A) + sin A - 2 sin^3 A

= 2 sin A - 2 sin^3 A + sin A - 2 sin^3 A

= 3 sin A - 4 sin^3 A    

Therefore, sin 3A = 3 sin A - 4 sin^3 A

4sin*3* A can be written as sin 60+a *sin 60 - a

Answer 2

please see the attachment