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For any triangle ABC, prove that a(cos C - cos B) = 2 (b-c) * cos² A/2.

For any triangle ABC, prove that
a(cos C - cos B) = 2 (b-c) * cos² A/2.

Grade:11

1 Answers

Arun
25750 Points
4 years ago
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]
 
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