Question

Prove that:

cos 20° cos 100° + cos 100° cos 140° − 140° cos 200° = −\[\frac{3}{4}\]

 

Answer 1

Consider LHS: 
\[ \cos 20^\circ \cos 100^\circ + \cos 100^\circ \cos 140^\circ - \cos 140^\circ \cos 200^\circ\]
\[ = \frac{1}{2}(2\cos 20^\circ \cos 100^\circ + 2\cos 100^\circ \cos 140^\circ - 2\cos 140^\circ \cos 200^\circ)\]
\[ = \frac{1}{2}\left[ \cos\left( 100^\circ + 20^\circ \right)\cos \left( 100^\circ - 20^\circ \right) + \cos \left( 140^\circ + 100^\circ \right)\cos \left( 140^\circ - 100^\circ \right) - \cos \left( 200^\circ + 140^\circ \right)\cos \left( 200^\circ - 140^\circ \right) \right]\]
\[ = \frac{1}{2}\left[ \cos120^\circ + \cos80^\circ + \cos240^\circ + \cos40^\circ - \cos340^\circ - \cos60^\circ \right]\]
\[ = \frac{1}{2}\left[ \cos120^\circ + \cos240^\circ - \cos60^\circ + \cos80^\circ + \cos40^\circ - \cos340^\circ \right]\]
\[ = \frac{1}{2}\left[ \left( - \frac{1}{2} - \frac{1}{2} - \frac{1}{2} \right) + \cos80^\circ + \cos40^\circ - \cos340^\circ \right]\]
\[ = \frac{1}{2}\left[ - \frac{3}{2} + \left\{ 2\cos\left( \frac{80^\circ + 40^\circ}{2} \right)\cos\left( \frac{80^\circ - 40^\circ}{2} \right) - \cos\left( 360^\circ - 20^\circ \right) \right\} \right]\]
\[ = \frac{1}{2}\left[ - \frac{3}{2} + \left\{ 2\cos60^\circ\cos20^\circ - \cos20^\circ \right\} \right]\]
\[ = \frac{1}{2}\left[ - \frac{3}{2} + \cos20^\circ - \cos20^\circ \right]\]
\[ = \frac{1}{2}\left[ - \frac{3}{2} \right]\]
\[ = - \frac{3}{4} = RHS\]
Hence, LHS = RHS