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Prove That: Cos π 12 − Sin π 12 = 1 √ 2 - Mathematics

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प्रश्न

Prove that:

\[\cos\frac{\pi}{12} - \sin\frac{\pi}{12} = \frac{1}{\sqrt{2}}\]

 

योग
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उत्तर

\[LHS = \cos \frac{\pi}{12} - \sin \frac{\pi}{12}\]
\[ = \cos \left( \frac{\pi}{2} - \frac{5\pi}{12} \right) - \sin \frac{\pi}{12}\]
\[ = \sin \left( \frac{5\pi}{12} \right) - \sin \frac{\pi}{12}\]
\[ = 2\sin \left( \frac{\frac{5\pi}{12} - \frac{\pi}{12}}{2} \right) \cos \left( \frac{\frac{5\pi}{12} + \frac{\pi}{12}}{2} \right) \left\{ \because \sin A - \sin B = 2\sin \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right) \right\}\]
\[ = 2\sin \left( \frac{\pi}{6} \right) \cos \left( \frac{\pi}{4} \right)\]
\[ = 2 \times \frac{1}{2} \times \frac{1}{\sqrt{2}}\]
\[ = \frac{1}{\sqrt{2}}\]
Hence, LHS = RHS.

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Transformation Formulae
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अध्याय 8: Transformation formulae - Exercise 8.2 [पृष्ठ १७]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 8 Transformation formulae
Exercise 8.2 | Q 3.6 | पृष्ठ १७

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