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

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प्रश्न

In triangle ABC, prove the following: 

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

 

उत्तर

Let 

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

Then, 

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

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

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Sine and Cosine Formulae and Their Applications
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पाठ 10: Sine and cosine formulae and their applications - Exercise 10.1 [पृष्ठ १३]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 10 Sine and cosine formulae and their applications
Exercise 10.1 | Q 17 | पृष्ठ १३

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