Question

4sina*sin(a+60°) *sin(a+120°)=sin3a​

Answer 1

Given:-

• 4 sinA × sin( A + 60) × sin( A + 120°) = sin3a

To Prove:-

• Lhs = Rhs

Solution:-

➳ 4sinA.sin(60+A).sin(120+A)

➳ 4sinA[sin(60+A).sin(180–60+A)]

➳ 4sinA[sin(60+A).sin{180-(60-A)}]

➳ 4sinA[sin(60+A).sin(60-A)]

➳ 4sinA[sin² (60) - sin² A]

➳ 4sinA[(3/4)-sin² A]

➳ 3sinA - 4sin³ A

➳ sin3A

∴ lhs = rhs

Hence, proved.

Answer 2

Answer:

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