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

In triangle ABC, prove the following:

$b \cos B + c \cos C = a \cos \left( B - C \right)$

#### Solution

Let

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Then,

Consider the LHS of the equation $b \cos B + c \cos C = a \cos \left( B - C \right)$

$LHS = b\cos B + c\cos C$
$= k\left( \sin B\cos B + \sin C\cos C \right)$
$= \frac{k}{2}\left( 2sinBcosB + 2sinCcosC \right)$
$= \frac{k}{2}\left( \sin2B + \sin2C \right) . . . \left( 1 \right)$
$RHS = a\cos\left( B - C \right)$
$= k\sin A\cos\left( B - C \right)$
$= \frac{k}{2}\left[ 2\sin A\cos\left( B - C \right) \right]$
$= \frac{k}{2}\left[ \sin\left( A + B - C \right) + \sin\left( A - B + C \right) \right] \left[ \because 2\sin A\cos B = \sin\left( A + B \right) + \sin\left( A - B \right) \right]$
$= \frac{k}{2}\left[ \sin\left( \pi - C - C \right) + \sin\left( \pi - B - B \right) \right] \left[ \because \sin\left( \pi - A \right) = \sin A, A + B + C = \pi \right]$
$= \frac{k}{2}\left( \sin2C + \sin2B \right)$
$= \frac{k}{2}\left( \sin2B + \sin2C \right) = LHS \left[ from\left( 1 \right) \right]$
$\text{ Hence proved } .$

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 17 | Page 13