IF A+B+C=Π prove that sin(3A)+sin(3B)+sin(3C)=-4cos(3A÷2)cos(3B÷2)cos(3C÷2)
1 answer:
Answer:
LHS of the given equation is:
sinθ+sin3θ+sin5θ+sin7θ=(sinθ+sin7θ)+(sin3θ+sin5θ)=2sin8θ2.cos6θ2+2sin8θ2.cos2θ2 [since sinC+sinD=2sinC+D2.cosC−D2=2sin4θ.cos3θ+2sin4θ.cosθ=2sin4θ.[cos3θ+cosθ] [since cosC+cosD=2cosC+D2.cosC−D2=2sin4θ.[2cos2θ.cosθ]=4cosθ.cos2θ.sin4θsinθ+sin3θ+sin5θ+sin7θ=(sinθ+sin7θ)+(sin3θ+sin5θ)=2sin8θ2.cos6θ2+2sin8θ2.cos2θ2 [since sinC+sinD=2sinC+D2.cosC-D2=2sin4θ.cos3θ+2sin4θ.cosθ=2sin4θ.[cos3θ+cosθ] [since cosC+cosD=2cosC+D2.cosC-D2=2sin4θ.[2cos2θ.cosθ]=4cosθ.cos2θ.sin4θ
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Step-by-step explanation:
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