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P Prove That: Cos ( 3 π 4 + X ) − Cos ( 3 π 4 − X ) = − √ 2 Sin X - Mathematics

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प्रश्न

Prove that:

\[\cos\left( \frac{3\pi}{4} + x \right) - \cos\left( \frac{3\pi}{4} - x \right) = - \sqrt{2} \sin x\]

 

योग
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उत्तर

Consider LHS:
\[\cos \left( \frac{3\pi}{4} + x \right) - \cos \left( \frac{3\pi}{4} - x \right)\]
\[ = - 2\sin\left\{ \frac{\left( \frac{3\pi}{4} + x \right) + \left( \frac{3\pi}{4} - x \right)}{2} \right\} \sin \left\{ \frac{\left( \frac{3\pi}{4} + x \right) - \left( \frac{3\pi}{4} - x \right)}{2} \right\} \left\{ \because \cos A - \cos B = - 2\sin \left( \frac{A + B}{2} \right) \sin\left( \frac{A - B}{2} \right) \right\}\]
\[ = - 2\sin\frac{3\pi}{4} \sin x\]
\[ = - 2\sin \left( \pi - \frac{\pi}{4} \right) \sin x\]
\[ = - 2\sin \frac{\pi}{4} \sin x\]
\[ = - \sqrt{2}\sin x\]

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Transformation Formulae
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Transformation formulae - Exercise 8.2 [पृष्ठ १८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 8 Transformation formulae
Exercise 8.2 | Q 4.1 | पृष्ठ १८

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