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Prove that cos⁡xcos⁡x2−cos⁡3xcos⁡9x2=sin⁡7xsin⁡8x

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प्रश्न

Prove that \[\cos x \cos \frac{x}{2} - \cos 3x \cos\frac{9x}{2} = \sin 7x \sin 8x\]
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उत्तर

\[\text{ LHS }= \cos x \cos\frac{x}{2} - \cos 3x \cos\frac{9x}{2}\]

\[ = \frac{1}{2}\left[ 2\cos x \cos\frac{x}{2} - 2\cos 3x \cos\frac{9x}{2} \right]\]

\[ = \frac{1}{2}\left[ \cos\left( x + \frac{x}{2} \right) + \cos\left( x - \frac{x}{2} \right) - \cos\left( 3x + \frac{9x}{2} \right) - \cos\left( 3x - \frac{9x}{2} \right) \right]\]

\[ = \frac{1}{2}\left[ \cos\frac{3x}{2} + \cos\frac{x}{2} - \cos\frac{15x}{2} - \cos\frac{3x}{2} \right]\]

\[ = \frac{1}{2}\left[ \cos\frac{x}{2} - \cos\frac{15x}{2} \right]\]

\[ = \frac{1}{2}\left[ - 2\sin\left( \frac{x + 15x}{4} \right)\sin\left( \frac{x - 15x}{4} \right) \right]\]

\[ = \frac{1}{2}\left[ - 2\sin\left( 4x \right)\sin\left( - \frac{7x}{2} \right) \right]\]

\[ = \sin\left( 4x \right)\sin\left( \frac{7x}{2} \right) = \text{ RHS }\]

Hence, LHS = RHS

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Notes

Disclaimer: The given question is incorrect. The correct question should be

\[\cos x \cos \frac{x}{2} - \cos 3x \cos\frac{9x}{2} = \sin 4x \sin \frac{7x}{2}\]
Transformation Formulae
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Transformation formulae - Exercise 8.2 [पृष्ठ १८]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 8 Transformation formulae
Exercise 8.2 | Q 6.7 | पृष्ठ १८

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