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Question
Prove that:
cos 10° cos 30° cos 50° cos 70° = \[\frac{3}{16}\]
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Solution
\[LHS = \cos 10^\circ \cos 30^\circ \cos 50^\circ \cos 70^\circ\]
\[ = \frac{1}{2} \left[ 2\cos 10^\circ \cos 50^\circ \right] \cos 30^\circ \cos 70^\circ\]
\[ = \frac{1}{2} \left[ \cos \left( 10^\circ + 50^\circ\right) + \cos \left( 10^\circ - 50^\circ \right) \right] \cos 30^\circ \cos 70^\circ \left\{ \because 2\cos A \cos B = \cos\left( A + B \right) - \cos \left( A - B \right) \right\}\]
\[ = \frac{1}{2} \left[ \cos 60^\circ + \cos \left( - 40^\circ \right) \right] \cos 30^\circ \cos 70^\circ\]
\[ = \frac{1}{2} \left[ \frac{1}{2} + \cos 40^\circ \right]\left( \frac{\sqrt{3}}{2} \right) \times \cos 70^\circ\]
\[= \frac{\sqrt{3}}{4}\cos 70^\circ\left[ \frac{1}{2} + \cos 40^\circ \right]\]
\[ = \frac{\sqrt{3}}{8}\cos 70^\circ + \frac{\sqrt{3}}{4}\left[ \cos 70^\circ \cos 40^\circ \right]\]
\[ = \frac{\sqrt{3}}{8}\cos 70^\circ + \frac{\sqrt{3}}{8}\left[ 2\cos 70^\circ \cos 40^\circ \right]\]
\[ = \frac{\sqrt{3}}{8}\cos 70^\circ + \frac{\sqrt{3}}{8}\left[ \cos \left( 70^\circ + 40^\circ \right) + \cos \left( 70^\circ - 40^\circ \right) \right]\]
\[ = \frac{\sqrt{3}}{8}\cos 70^\circ + \frac{\sqrt{3}}{8}\left[ \cos 110^\circ + \cos 30^\circ \right]\]
\[ = \frac{\sqrt{3}}{8}\cos 70^\circ + \frac{\sqrt{3}}{8}\left[ \cos \left( 180^\circ - 70^\circ \right) + \frac{\sqrt{3}}{2} \right]\]
\[ = \frac{\sqrt{3}}{2}\cos 70^\circ - \frac{\sqrt{3}}{8}\cos 70^\circ + \frac{3}{16} \left[ \because \cos \left( 180^\circ - 70^\circ \right) = - \cos 70^\circ \right]\]
\[ = \frac{3}{16} = RHS\]
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