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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)

Grade:11

1 Answers

Vikas TU
14149 Points
4 years ago
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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