If A+B+C=Π then PT sin(B+2C) +sin(C+2A) +sin(A+2B)=(4sin(B-C)/2)(Sin(C-A)/2)(sin(A-B)/2)

Question

Vikas TU · Answer

Given that A+B+C=1801 LHS we can write it as: sin(B+2C) +sin(C+2A) +sin(A+2B)=Sin(180-(A-C))+ Sin(180-(C-B))+ Sin(180-(B-A)) =Sin(A-C)+Sin(C-B)+Sin(B-A) Manipulating again we can write: =2Sin((C-B)/2)cos((C-B)/2)+2Sin((B-C)/2)Cos((2A-B-C)/2) =2Sin((B-C)/2)[-cos((C-B)/2)+ Cos((2A-B-C)/2)] =(4sin(B-C)/2)(Sin(C-A)/2)(sin(A-B)/2) =RHS

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If A+B+C=Π then PT sin(B+2C) +sin(C+2A) +sin(A+2B)=(4sin(B-C)/2)(Sin(C-A)/2)(sin(A-B)/2)

If A+B+C=Π then PT sin(B+2C) +sin(C+2A) +sin(A+2B)=(4sin(B-C)/2)(Sin(C-A)/2)(sin(A-B)/2)

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