Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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Prove That: Sin 8 π 3 Cos 23 π 6 + Cos 13 π 3 Sin 35 π 6 = 1 2 - Mathematics

Prove that:

\[\sin\frac{8\pi}{3}\cos\frac{23\pi}{6} + \cos\frac{13\pi}{3}\sin\frac{35\pi}{6} = \frac{1}{2}\]

 

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Solution

 LHS = \[\sin\frac{8\pi}{3}\cos\frac{23\pi}{6} + \cos\frac{13\pi}{3}\sin\frac{35\pi}{6}\]
\[ = \sin \left( \frac{8}{3} \times 180^\circ \right) \cos \left( \frac{23}{6} \times 180^\circ \right) + \cos\left( \frac{13}{3} \times 180^\circ \right)\sin\left( \frac{35}{6} \times 180^\circ \right)\]
\[ = \sin \left( 480^\circ \right) \cos \left( 690^\circ \right) + \cos \left( 780^\circ \right) \sin \left( 1050^\circ \right)\]
\[ = \sin \left( 90^\circ \times 5 + 30^\circ \right) \cos \left( 90^\circ \times 7 + 60^\circ \right) + \cos \left( 90^\circ \times 8 + 60^\circ \right)\sin \left( 90^\circ \times 11 + 60^\circ \right)\]
\[ = \cos \left( 30^\circ \right) \sin \left( 60^\circ \right) + \cos \left( 60^\circ \right)\left[ - \cos \left( 60^\circ \right) \right]\]
\[ = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \left( - \frac{1}{2} \right)\]
\[ = \frac{3}{4} - \frac{1}{4}\]
\[ = \frac{2}{4}\]
\[ = \frac{1}{2}\]
= RHS
Hence proved .

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 5 Trigonometric Functions
Exercise 5.3 | Q 2.2 | Page 39
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