1
Question
∑cos2αsinβsinγsin(β−γ)=−sin(β−γ)sin(γ−α)sin(α−β).
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Solution
T1=(1+cos2α2)⋅12[cos(β−γ)−cos(β+γ)]sin(β−γ)
∴ L.H.S. =18(1+cos2α)[sin2(β−γ)−{sin2β−sin2γ}]
=18[∑sin2(β−γ)−∑(sin2β−sin2γ)+∑cos2αsin2(β−γ)−∑cos2α(sin2β−sin2γ)]
The second sigma is clearly zero.
The third sigma is of 6 terms which cancel in pairs.
∑cos2α{sin2βcos2γ−cos2βsin2γ}=0
The last sigma is of six terms and taken in pairs it is {sin2(β−γ)+sin2(γ−α)+sin2(α−β)}=∑sin2(β−γ)
∴ L.H.S. =18⋅2⋅∑sin2(β−γ)
=14{−4sin(β−γ)sin(γ−α)sin(α−β)}
=−sin(β−γ)sin(γ−α)sin(α−β) as in part (a).
∴ L.H.S. =18(1+cos2α)[sin2(β−γ)−{sin2β−sin2γ}]
=18[∑sin2(β−γ)−∑(sin2β−sin2γ)+∑cos2αsin2(β−γ)−∑cos2α(sin2β−sin2γ)]
The second sigma is clearly zero.
The third sigma is of 6 terms which cancel in pairs.
∑cos2α{sin2βcos2γ−cos2βsin2γ}=0
The last sigma is of six terms and taken in pairs it is {sin2(β−γ)+sin2(γ−α)+sin2(α−β)}=∑sin2(β−γ)
∴ L.H.S. =18⋅2⋅∑sin2(β−γ)
=14{−4sin(β−γ)sin(γ−α)sin(α−β)}
=−sin(β−γ)sin(γ−α)sin(α−β) as in part (a).
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