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Prove That:Cos 40° Cos 80° Cos 160° = \[- \Frac{1}{8}\] - Mathematics

Sum

Prove that:
cos 40° cos 80° cos 160° = \[- \frac{1}{8}\]

 

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Solution

\[LHS = \cos 40^\circ \cos 80^\circ \cos 160^\circ\]
\[ = \frac{1}{2}\left[ 2\cos 40^\circ \cos 80^\circ \right] \cos 160^\circ\]
\[ = \frac{1}{2}\left[ \cos \left( 40^\circ + 80^\circ \right) + \cos \left( 40^\circ - 80^\circ \right) \right] \cos 160^\circ\]
\[ = \frac{1}{2}\left[ \cos 120^\circ + \cos \left( - 40^\circ \right) \right] \cos 160^\circ\]
\[ = \frac{1}{2}\cos \left( 160^\circ \right)\left[ - \frac{1}{2} + \cos 40^\circ \right]\]
\[ = - \frac{1}{4}\cos 160^\circ + \frac{1}{2}\cos 160^\circ \cos 40^\circ\]
\[= - \frac{1}{4}\cos 160^\circ + \frac{1}{4}\left[ 2\cos 160^\circ \cos 40^\circ \right]\]
\[ = - \frac{1}{4}\cos 160^\circ + \frac{1}{4}\left[ \cos \left( 160^\circ + 40^\circ \right) + \cos \left( 160^\circ - 40^\circ \right) \right]\]
\[ = - \frac{1}{4}\cos 160^\circ + \frac{1}{4}\left[ \cos 200^\circ + \cos 120^\circ \right]\]
\[ = - \frac{1}{4}\cos 160^\circ + \frac{1}{4}\left[ \cos \left( 360^\circ - 160^\circ \right) - \frac{1}{2} \right]\]
\[ = - \frac{1}{4}\cos 160^\circ + \frac{1}{4}\cos 160^\circ - \frac{1}{8} \left[ \because \cos \left( 360^\circ - 160^\circ \right) = \cos 160^\circ \right]\]
\[ = - \frac{1}{8} = RHS\]

 

Concept: Transformation Formulae
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 8 Transformation formulae
Exercise 8.1 | Q 5.2 | Page 7
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