Prove that : a (cos C − cos B) = 2 (b − C) cos2\(\frac{A}{2}.\)

1.	Prove that : a (cos C − cos B) = 2 (b − C) cos2\(\frac{A}{2}.\)
Answer» LHS = a [cos C – cos B] = \(a[2\,sin{\frac{B+C}{2}sin\frac{B-C}{2}}]\) \(= a[2\,cos{\frac{A}{2}sin\frac{B-C}{2}}]\) \(= 2k\,sin\,A[cos{\frac{A}{2}sin\frac{B-C}{2}}]\) \(= 2k\,2\,sin\,\frac{A}{2}\,cos\frac{A}{2}[cos{\frac{A}{2}sin\frac{B-C}{2}}]\) \(= 2k\,cos^2\frac{A}{2}[2\,sin{\frac{A}{2}sin\frac{B-C}{2}}]\) \(= 2k\,cos^2\frac{A}{2}k[sin\{90°-(\frac{B+C}{2})\}sin\frac{B-C}{2}]\) \(= 2\,cos^2\frac{A}{2}k[cos(\frac{B+C}{2})sin(\frac{B-C}{2})]\) \(= 2\,cos^2\frac{A}{2}k[sin\,B\,-sin\,C]\) \(= 2\,cos^2\frac{A}{2}[k\,sin\,B\,-k\,sin\,C]\) \(= 2\,cos^2\frac{A}{2}[b \,-\,c]\) = RHS ∴ LHS = RHS

Answer»

LHS = a [cos C – cos B]

= \(a[2\,sin{\frac{B+C}{2}sin\frac{B-C}{2}}]\)

\(= a[2\,cos{\frac{A}{2}sin\frac{B-C}{2}}]\)

\(= 2k\,sin\,A[cos{\frac{A}{2}sin\frac{B-C}{2}}]\)

\(= 2k\,2\,sin\,\frac{A}{2}\,cos\frac{A}{2}[cos{\frac{A}{2}sin\frac{B-C}{2}}]\)

\(= 2k\,cos^2\frac{A}{2}[2\,sin{\frac{A}{2}sin\frac{B-C}{2}}]\)

\(= 2k\,cos^2\frac{A}{2}k[sin\{90°-(\frac{B+C}{2})\}sin\frac{B-C}{2}]\)

\(= 2\,cos^2\frac{A}{2}k[cos(\frac{B+C}{2})sin(\frac{B-C}{2})]\)

\(= 2\,cos^2\frac{A}{2}k[sin\,B\,-sin\,C]\)

\(= 2\,cos^2\frac{A}{2}[k\,sin\,B\,-k\,sin\,C]\)

\(= 2\,cos^2\frac{A}{2}[b \,-\,c]\) = RHS

∴ LHS = RHS

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