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In triangle ABC, prove the following: a cos A + b cos B + c cos C = 2 b sin A sin C = 2 c sin A sin B

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प्रश्न

In triangle ABC, prove the following: 

\[a \cos A + b\cos B + c \cos C = 2b \sin A \sin C = 2 c \sin A \sin B\]

 

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उत्तर

\[Let\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k . . . \left( 1 \right)\]
\[\text{ Consider the LHS of the equation a }\cos A + b\cos B + c \cos C . \]
\[a\cos A + b\cos B + c\cos C = k\left( \sin A\cos A + \sin B\cos B + \sin C\cos C \right) \]
\[ = \frac{k}{2}\left( 2sinAcosA + 2sinAcosA + 2sinCcosC \right)\]
\[ = \frac{k}{2}\left( \sin2A + \sin2 B + \sin2 C \right)\]
\[ = \frac{k}{2}\left[ 2\sin\left( A + B \right)\cos\left( A - B \right) + 2\sin C\cos C \right]\]
\[ = \frac{k}{2}\left[ 2\sin\left( \pi - C \right)\cos\left( A - B \right) + 2\sin C\cos C \right]\]
\[ = \frac{k}{2}\left[ 2\sin C\cos\left( A - B \right) + 2\sin C\cos C \right]\]
\[ = \frac{2k\sin C}{2}\left[ \cos\left( A - B \right) + \cos C \right] \]

\[= k\sin C\left[ \cos\left( A - B \right) + \cos\left\{ \pi - \left( A + B \right) \right\} \right]\]
\[ = k\sin C\left[ \cos\left( A - B \right) - \cos\left( A + B \right) \right]\]
\[ = k\sin C\left[ 2\sin Asin B \right]\]
\[ = 2k\sin A\sin B\sin C . . . (1)\]

\[\text{ Now }, \]
\[\text{ on putting } k\sin C = \text{ C in equation } (1), \text{ we get }: \]
\[2c\sin A\sin B\]
\[\text{ and on putting k }\sin B = \text{ b in equation } (1), \text{ we get }: \]
\[2b\sin A\sin C\]

So, from (1), we have 

\[a \cos A + b\cos B + c \cos C = 2b \sin A \sin C = 2 c \sin A \sin B\]

Hence proved.

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Sine and Cosine Formulae and Their Applications
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 10: Sine and cosine formulae and their applications - Exercise 10.1 [पृष्ठ १३]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 10 Sine and cosine formulae and their applications
Exercise 10.1 | Q 22 | पृष्ठ १३

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