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Question
Prove that:
cos 80° + cos 40° − cos 20° = 0
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Solution
Consider LHS:
\[\cos 80^\circ + \cos 40^\circ - \cos 20^\circ\]
\[ = 2\cos \left( \frac{80^\circ + 40^\circ}{2} \right) \cos \left( \frac{80^\circ - 40^\circ}{2} \right) - \cos 20^\circ \left\{ \because \cos A + \cos B = 2\cos\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) \right\}\]
\[ = 2\cos 60^\circ \cos 20^\circ - \cos 20^\circ\]
\[ = 2 \times \frac{1}{2}\cos 20^\circ - \cos20^\circ\]
\[ = \cos 20^\circ - \cos 20^\circ\]
\[ = 0\]
Hence, LHS = RHS.
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