मराठी

Show That: Sin A Sin (B − C) + Sin B Sin (C − A) + Sin C Sin (A − B) = 0 - Mathematics

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प्रश्न

Show that:
sin A sin (B − C) + sin B sin (C − A) + sin C sin (A − B) = 0

बेरीज
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उत्तर

Consider LHS: 
\[\sin A \sin \left( B - C \right) + \sin B \sin \left( C - A \right) + \sin C \sin \left( A - B \right)\]
\[ = \frac{1}{2}\left[ 2\sin A \sin \left( B - C \right) \right] + \frac{1}{2}\left[ 2\sin B \sin \left( C - A \right) \right] + \frac{1}{2}\left[ 2\sin C \sin \left( A - B \right) \right]\]
\[ = \frac{1}{2}\left[ \cos \left\{ A - \left( B - C \right) \right\} - \cos \left\{ A + \left( B - C \right) \right\} \right] + \frac{1}{2}\left[ \cos \left\{ B - \left( C - A \right) \right\} - \cos \left\{ B + \left( C - A \right) \right\} \right] + \frac{1}{2}\left[ \cos \left\{ C - \left( A - B \right) \right\} - \cos \left\{ C + \left( A - B \right) \right\} \right]\]
\[ = \frac{1}{2}\left[ \cos \left( A - B + C \right) - \cos \left( A + B - C \right) \right] + \frac{1}{2}\left[ \cos \left( B - C + A \right) - \cos\left( B + C - A \right) \right] + \frac{1}{2}\left[ \cos\left( C - A + B \right) - \cos\left( C + A - B \right) \right]\]
\[ = \frac{1}{2}\cos\left( A - B + C \right) - \frac{1}{2}\cos \left( A + B - C \right) + \frac{1}{2}\cos \left( B - C + A \right) - \frac{1}{2}\cos \left( B + C - A \right) + \frac{1}{2}\cos \left( C - A + B \right) - \frac{1}{2}\cos\left( C + A - B \right)\]
\[ = \frac{1}{2}\cos\left( A - B + C \right) - \frac{1}{2}\cos\left( A + B - C \right) + \frac{1}{2}\cos\left( A + B - C \right) - \frac{1}{2}\cos\left( B + C - A \right) + \frac{1}{2}\cos\left( B + C - A \right) - \frac{1}{2}\cos\left( A - B + C \right)\]
\[ = 0\]
 = RHS

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Transformation Formulae
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Transformation formulae - Exercise 8.1 [पृष्ठ ७]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 8 Transformation formulae
Exercise 8.1 | Q 6.1 | पृष्ठ ७

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