मराठी

Prove That: Sin 65 ∘ + Cos 65 ∘ = √ 2 Cos 20 ∘

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प्रश्न

Prove that:

\[\sin 65^\circ + \cos 65^\circ = \sqrt{2} \cos 20^\circ\]
बेरीज
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उत्तर

Consider LHS:
\[\sin 65^\circ + \cos 65^\circ\]
\[ = \sin 65^\circ + \cos \left( 90^\circ - 25^\circ \right)\]
\[ = \sin 65^\circ + \sin 25^\circ\]
\[ = 2\sin \left( \frac{65^\circ + 25^\circ}{2} \right) \cos \left( \frac{65^\circ - 25^\circ}{2} \right) \left\{ \because \sin A + \sin B = 2\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \right\}\]
\[ = 2\sin 45^\circ \cos 20^\circ\]
\[ = 2 \times \frac{1}{\sqrt{2}} \cos 20^\circ\]
\[ = \sqrt{2}\cos 20^\circ\]
 = 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 5.1 | पृष्ठ १८

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