Question

If A+B+C = 180, Then prove that
Sin(B + 2C) + Sin(C+2A)+ Sin(A+2B)=4Sin(B-C/2)Sin(C-A/2)Cos(A-B/2)

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Answer 1

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

Step-by-step explanation:

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