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Question

Prove the following 2sinA·cos3A-2sin3A·cosA=sin4A2


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Solution

Proof:

Taking LHS,

LHS=2sinA·cos3A-2sin3A·cosA=2sinA·cosA(cos2A-sin2A)=sin2A·cos2A2sinA·cosA=sin2Aandcos2A-sin2A=cos2A=2sin2A·cos2A2=sin(2·2A)22sinA·cosA=sin2A=sin4A2=RHS

Hence, proved.


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