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Question

If A+B+C=π, then prove that sinA2+sinB2+sinC21=4sinπ44.sinπB4.sinπC4

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Solution

A+B+C=πA+B=πC (1)B+C=πA (2)A+C=πB (3)
To Prove: sinA2+sinB2+sinC21=4sinπ44.sinπB4.sinπC4LHS=sinA2+sinB2+sinC21=sinA2+sinB2+sinC2sinπ2 (1=sinπ2)=2sinA2+B22cosA2B22+2cosC2+π22sinC2π22=2sinA+B4cosAB4+2cosC+π4sinCπ4=2sinπC4cosAB4+2cosC+π4(sinπC4)=2sinπC4[cosAB4cosC+π4]=2sinπC4[2sinAB4+C+π42sinAB4C+π2]=4sinπC4[sin(A+C)B+π8sinA+π(B+C)8]=4sinπC4[sin(A+C)B+π8sinA+π(π4)8]
=4sinπC4[sin2π2B8sin2A2π8]=4sinπC4[sinπB4sinAπ4]=4sinπC4 sinπB4 4sinπA4=RHS
Hence Proved.

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