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Prove That: Sin 80° − Cos 70° = Cos 50°

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Question

Prove that:

sin 80° − cos 70° = cos 50°
Sum
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Solution

Consider LHS: 
\[\sin 80^\circ - \cos 70^\circ\]
\[ = \sin 80^\circ - \cos \left( 90^\circ - 20^\circ \right)\]
\[ = \sin 80^\circ - \sin 20^\circ\]
\[ = 2\sin \left( \frac{80^\circ - 20^\circ}{2} \right) \cos \left( \frac{80^\circ + 20^\circ}{2} \right) \left\{ \because \sin A - \sin B = 2\sin \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right) \right\}\]
\[ = 2\sin 30^\circ \cos 50^\circ\]
\[ = 2 \times \frac{1}{2}\cos 50^\circ\]
\[ = \cos 50^\circ\]
= RHS
Hence, LHS = RHS.

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Transformation Formulae
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Chapter 8: Transformation formulae - Exercise 8.2 [Page 17]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 8 Transformation formulae
Exercise 8.2 | Q 3.7 | Page 17

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