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In Triangle Abc, Prove the Following: ( a − B ) Cos C 2 = C Sin ( a − B 2 )

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Question

In triangle ABC, prove the following: 

\[\left( a - b \right) \cos \frac{C}{2} = c \sin \left( \frac{A - B}{2} \right)\]

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Solution

Let

\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\]                                  ...(1) 
Consider the LHS of the equation
\[\left( a - b \right) \cos \frac{C}{2} = c \sin \left( \frac{A - B}{2} \right)\] 
\[LHS = \left( a - b \right)\cos\frac{C}{2}\]
\[ = k\left( \sin A - \sin B \right)\cos\frac{C}{2} \left( \text{ using }\left( 1 \right) \right) \]
\[ = k \times 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right)\cos\frac{C}{2}\]
\[ = 2$k\sin\left( \frac{A - B}{2} \right)\cos$\left( \frac{A + B}{2} \right)\cos\left( \frac{\pi - \left( A + B \right)}{2} \right) \left[ \because A + B + C = \pi \right]\]
\[ = 2k\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right)\sin\left( \frac{A + B}{2} \right)\]
\[ = k\sin\left( \frac{A - B}{2} \right)\sin\left( A + B \right) \left[ \because 2\cos\left( \frac{A + B}{2} \right)\sin\left( \frac{A + B}{2} \right) = \sin\left( A + B \right) \right]\]
\[ = k\sin\frac{A - B}{2}\sin\left( \pi - C \right) \left[ \because A + B + C = \pi \right]\]
\[ = k\sin C\sin\left( \frac{A - B}{2} \right) \]
\[ = C\sin\left( \frac{A - B}{2} \right) = RHS \]
Hence proved.
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Sine and Cosine Formulae and Their Applications
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Chapter 10: Sine and cosine formulae and their applications - Exercise 10.1 [Page 13]

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R.D. Sharma Mathematics [English] Class 11
Chapter 10 Sine and cosine formulae and their applications
Exercise 10.1 | Q 5 | Page 13

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