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

In triangle ABC, prove the following:

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

#### Solution

Let

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$                                  ...(1)
Consider the LHS of the equation
$\left( a - b \right) \cos \frac{C}{2} = c \sin \left( \frac{A - B}{2} \right)$
$LHS = \left( a - b \right)\cos\frac{C}{2}$
$= k\left( \sin A - \sin B \right)\cos\frac{C}{2} \left( \text{ using }\left( 1 \right) \right)$
$= k \times 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right)\cos\frac{C}{2}$
$= 2k\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right)\cos\left( \frac{\pi - \left( A + B \right)}{2} \right) \left[ \because A + B + C = \pi \right]$
$= 2k\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right)\sin\left( \frac{A + B}{2} \right)$
$= k\sin\left( \frac{A - B}{2} \right)\sin\left( A + B \right) \left[ \because 2\cos\left( \frac{A + B}{2} \right)\sin\left( \frac{A + B}{2} \right) = \sin\left( A + B \right) \right]$
$= k\sin\frac{A - B}{2}\sin\left( \pi - C \right) \left[ \because A + B + C = \pi \right]$
$= k\sin C\sin\left( \frac{A - B}{2} \right)$
$= C\sin\left( \frac{A - B}{2} \right) = RHS$
Hence proved.
Concept: Sine and Cosine Formulae and Their Applications
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 10 Sine and cosine formulae and their applications
Exercise 10.1 | Q 5 | Page 13