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Question
Prove that:
\[\tan 4\pi - \cos\frac{3\pi}{2} - \sin\frac{5\pi}{6}\cos\frac{2\pi}{3} = \frac{1}{4}\]
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Solution
\[ 4\pi = 720^\circ, \frac{3\pi}{2} = 270^\circ, \frac{5\pi}{6} = 150^\circ, \frac{2\pi}{3} = 120^\circ\]
LHS = \[\tan\left( 720^\circ \right) - \cos\left( 270^\circ \right) - \sin\left( 150^\circ \right) \cos\left( 120^\circ \right)\]
\[ = \tan\left( 90^\circ \times 8 + 0^\circ \right) - \cos\left( 90^\circ \times 3 + 0^\circ \right) - \sin\left( 90^\circ \times 1 + 60^\circ \right) \cos\left( 90^\circ \times 1 + 30^\circ \right)\]
\[ = \tan\left( 0^\circ \right) - \sin\left( 0^\circ \right) - \cos\left( 60^\circ \right) \left[ - \sin\left( 30^\circ \right) \right]\]
\[ = \tan\left( 0^\circ \right) - \sin\left( 0^\circ \right) + \cos\left( 60^\circ \right) \sin\left( 30^\circ \right)\]
\[ = 0 - 0 + \frac{1}{2} \times \frac{1}{2}\]
\[ = \frac{1}{4}\]
= RHS
Hence proved.
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