मराठी

Prove That: Sin X 2 Sin 7 X 2 + Sin 3 X 2 Sin 11 X 2 = Sin 2 X Sin 5 X .

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प्रश्न

Prove that:
\[\sin\frac{x}{2}\sin\frac{7x}{2} + \sin\frac{3x}{2}\sin\frac{11x}{2} = \sin 2x \sin 5x .\]

 

बेरीज
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उत्तर

\[\text{ LHS }= \sin\frac{x}{2}\sin\frac{7x}{2} + \sin\frac{3x}{2}\sin\frac{11x}{2}\]
\[ = \frac{1}{2}\left[ 2\sin\frac{x}{2}\sin\frac{7x}{2} + 2\sin\frac{3x}{2}\sin\frac{11x}{2} \right]\]
\[ = \frac{1}{2}\left[ \cos\left( \frac{7x}{2} - \frac{x}{2} \right) - \cos\left( \frac{7x}{2} + \frac{x}{2} \right) + \cos\left( \frac{11x}{2} - \frac{3x}{2} \right) - \cos\left( \frac{11x}{2} + \frac{3x}{2} \right) \right]\]
\[ = \frac{1}{2}\left[ \cos3x - \cos4x + \cos4x - \cos7x \right]\]
\[ = \frac{1}{2}\left[ \cos3x - \cos7x \right]\]
\[ = \frac{1}{2}\left[ - 2\sin\left( \frac{3x + 7x}{2} \right)\sin\left( \frac{3x - 7x}{2} \right) \right]\]
\[ = \frac{1}{2}\left[ - 2\sin\left( 5x \right)\sin\left( - 2x \right) \right]\]
\[ = \sin\left( 5x \right)\sin\left( 2x \right) =\text{ RHS }\] 
Hence, LHS = RHS

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Transformation Formulae
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पाठ 8: Transformation formulae - Exercise 8.2 [पृष्ठ १८]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 8 Transformation formulae
Exercise 8.2 | Q 6.6 | पृष्ठ १८

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