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prove that a(cosC-cosB)=2(b-c)cos^2(A/2)

prove that a(cosC-cosB)=2(b-c)cos^2(A/2)
 

Grade:11

2 Answers

Nishant Vora IIT Patna
askIITians Faculty 2467 Points
6 years ago
Hello student, Please find the answer to your question below

a/ 2(b-c) = cos^2(A/2)/ (cosC-cosB)
lhs = a / 2 (b – c )
Now use sine rule
a/ sinA = b/ sinB = c/ sinC = k
So, lhs = a / 2 (b – c )= sinA / 2(sinB -sinC)
=SinA /2* 2 cos(B+C/2) cos (B-C/2)
= sinA / 4 cos (pi/2 – A/2) cos (B-C/2)
=2 sin(A/2) cos (A/2) / 4 sin(A/2) cos (B-C/2)
= cos (A/2) / cos(B-C/2)
=cos2(A/2)/ Cos(A/2) cos(B-C/2)
= cos2(A/2)/ Cos(90 – (B+C)/2) cos(B-C/2)
=2 cos2(A/2)/ 2 sin((B+C)/2) cos(B-C/2)
= 2 cos2(A/2)/ 2 sin((B+C)/2) cos(B-C/2)
=cos2(A/2)/ (cosC-cosB)
ankit singh
askIITians Faculty 614 Points
11 months ago
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
 

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