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`        For any triangle ABC, prove thata (cos C - cosB) = 2 (b-c) * cos² A/2.`
6 months ago

```							Dear student i) By the application of sum/product rule of trigonometric relation, cos(C) - cos(B) = 2sin{(B+C)/2}*sin{(B-C)/2} ii) By angle sum property of a triangle, A + B + C = 180 deg ==> (B+C)/2 = 90 - (A/2) ==> sin{(B+C)/2} = sin{90 - (A/2)} = cos(A/2) iii) From sine law of triangle, a/sin(A) = b/sin(B) = c/sin(C);let each of them equal to k. ==> a = ksin(A) = 2ksin(A/2)*cos(A/2) [Application of multiple angle identity] iv) Thus from the above 3 steps, a{cos(C) - cos(B)} = k*2sin(A/2)cos(A/2)*2cos(A/2)sin{(B-C)/2... = k*{2cos²(A/2)}*[2sin(A/2)sin{(B-C)/2}] =k*{2cos²(A/2)}*[2cos{(B+C)/2}sin{(B-C...  [From step (ii) above, also cos(B+C)/2 = cos(90 - A/2) = sin(A/2)] = k*{2cos²(A/2)}*[sin(B) - sin(C)] [Application of sum/product rule] = {2cos²(A/2)}[ksin(B) - ksin(C)] = 2cos²(A/2)(b - c) [By sine law of triangle, ksin(B) = b and ksin(C) = c] Thus a{cos(C) - cos(B)} = 2(b - c)cos²(A/2) [Proved] Hope it helps RegardsArun (askIITians forum expert)
```
6 months ago
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