Question

Please answer fast I will mark you as the brainliest

Prove that :
a(cos C - cosB) = 2(b-c ) cos^A/2

Answer 1

Answer:

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)

hope it will help u

check it should be cos^2a/2