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

Explanation

From sine law, we have,

`a/sinA = b/sinB = c/sinC = k`

`=> a = ksinA, b = ksin B, c = ksinC`

Now, `L.H.S. = a(cosC-cosB) = ksinA(cosC-cosB)`

`=k(2sin(A/2)cos(A/2))[2sin((B+C)/2)sin((B-C)/2)]`

`=k(2sin(A/2)cos(A/2))[2sin((pi-A)/2)sin((B-C)/2)]...[As A+B+C = pi]`

`=4ksin((pi-(B+C))/2)cos(A/2)cos(A/2)sin((B-C)/2)`

`=4kcos^2(A/2)cos((B+C)/2)sin((B-C)/2)`

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