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Question

sin(BC2)=bca cos A2

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Solution

Let a=k sin A, b=k sin B, c=k sin CRHSbcacosA2=k sin Bk sin Ck sin A.cosA2=sin Bsin Csin A.cosA2=2cosB+C2.sinBC22sinA2.cosA2=cosπA2sinBC2sinA2=sinA2sinBC2sinA2=sinBC2=RHS


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