मराठी

Show that : Sin 25 ∘ Cos 115 ∘ = 1 2 ( Sin 140 ∘ − 1 )

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प्रश्न

Show that :

\[\sin 25^\circ \cos 115^\circ = \frac{1}{2}\left( \sin 140^\circ - 1 \right)\]
बेरीज
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उत्तर

\[LHS = 2\sin25^\circ \cos 115^\circ\]
\[ = \frac{\sin \left( 25^\circ + 115^\circ \right) + \sin \left( 25^\circ - 115^\circ \right)}{2} \left[ \because \sin A \cos B = \frac{1}{2}\left\{ \sin (A + B) + \sin (A - B) \right\} \right]\]
\[ = \frac{\sin 140^\circ + \sin \left( - 90^\circ \right)}{2}\]
\[ = \frac{\sin 140^\circ - \sin \left( 90^\circ \right)}{2}\]
\[ = \frac{\sin 140^\circ - 1}{2} \]
\[RHS = \frac{\sin 140^\circ - 1}{2}\]
Hence, LHS = RHS

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Transformation Formulae
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पाठ 8: Transformation formulae - Exercise 8.1 [पृष्ठ ६]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 8 Transformation formulae
Exercise 8.1 | Q 3.2 | पृष्ठ ६

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