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Prove this trignomeric solutionSin(B-C)/2 =(b-c)/(b+c)*CosA/2

saiyam , 5 Years ago
Grade 11
anser 1 Answers
Arun

Last Activity: 5 Years ago

 
Let a/sin A = b/sin B = c/sinC = k
Then, 
a = k sinA, b = k sinB, c = k sinC
RHS
b-c/a cos A/2
= (ksinB - ksinC/k sinA )cos A/2
= {[2 cos B+C/2 Sin B-C/2]/ Sin A} cos A/2
=[2* Sin B-C/2 cos (pi/2-A/2) .cos A/2] / sinA
= [SinA* sinB-C/2] / sinA
= Sin B-C/2
= LHS
 
Regards
Arun

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