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A Cos a + B Cos B + C Cos C = 2b Sin a Sin C - Mathematics

a cos + b cos B + c cos C = 2sin sin 

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Solution

\[\text{ By sine rule, we know that }\]

\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \left( say \right)\]

\[ \Rightarrow a = k \sin A, b = k \sin B, c = k \sin C\]

\[\text{ Now }, \]

\[LHS = a \cos A + b \cos B + c \cos C\]

\[ = k \sin A \cos A + k \sin B \cos B + k \sin C \cos C\]

\[ = \frac{k}{2} \left( 2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C \right)\]

\[ = \frac{k}{2} \left( \sin 2A + \sin 2B + 2 \sin C \cos C \right)\]

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

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

\[ = \frac{k}{2} \left( 2 \sin \left( \pi - C \right) \cos \left( A - B \right) + 2 \sin C \cos C \right) \left( \because A + B + C = \pi \right)\]

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

\[ = \frac{k}{2} \times 2 \sin C\left( \cos \left( A - B \right) + \cos C \right)\]

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

\[ = k \sin C\left( 2 \cos \left( \frac{\pi - B - B}{2} \right)\cos \left( \frac{B + C - A}{2} \right) \right) \left( \because A + B + C = \pi \right)\]

\[ = k \sin C\left( 2 \cos \left( \frac{\pi - 2B}{2} \right)\cos \left( \frac{\pi - 2A}{2} \right) \right) \left( \because A + B + C = \pi \right)\]

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

\[ = 2k \sin C\left( \sin B \sin A \right)\]

\[ = 2 \left( k \sin B \right) \sin A \sin C\]

\[ = 2b \sin A \sin C\]

\[ = RHS\]

\[ \therefore LHS = RHS\]

Hence, a cos + b cos B + c cos C = 2sin sin C.

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 10.2 | Q 11 | Page 25
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