# Prove That:Cos 570° Sin 510° + Sin (−330°) Cos (−390°) = 0 - Mathematics

Prove that:cos 570° sin 510° + sin (−330°) cos (−390°) = 0

#### Solution

LHS =$\cos \left( 570^\circ \right)\sin \left( 510^\circ \right) + \sin \left( - 330^\circ \right)\cos \left( - 390^\circ \right)$
$= \cos \left( 570^\circ \right) \sin \left( 510^\circ \right) + \left[ - \sin \left( 330^\circ \right) \right]\cos \left( 390^\circ \right) \left[ \because \sin\left( - x \right) = - \sin x and \cos\left( - x \right) = \cos x \right]$
$= \cos \left( 570^\circ \right)\sin\left( 510^\circ \right) - \sin \left( 330^\circ\right)$
$= \cos \left( 90^\circ \times 6 + 30^\circ \right) \sin \left( 90^\circ \times 5 + 60^\circ \right) - \sin \left( 90^\circ \times 3 + 60^\circ \right) \cos \left( 90^\circ \times 4 + 30^\circ \right)$
$= - \cos \left( 30^\circ \right) \cos \left( 60^\circ \right) - \left[ - \cos \left( 60^\circ \right) \right] \cos \left( 30^\circ \right)$
$= - \cos \left( 30^\circ \right) \cos \left( 60^\circ \right) + \cos \left( 30^\circ \right) \sin \left( 60^\circ \right)$
$= 0$
= RHS
Hence proved .

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 5 Trigonometric Functions
Exercise 5.3 | Q 2.5 | Page 39