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Question
Show that `(sin 8x cos x - sin 6x cos 3x)/(cos 2x cos x - sin 3x sin 4x)` = tan 2x
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Solution
`(sin 8x cos x - sin 6x cos 3x)/(cos 2x cos x - sin 3x sin 4x) = (1/2 [sin 9x + sin 7x] - 1/2 [sin 9x + sin 3x])/(1/2 [cos 3x + cos x] - 1/2 [cos x - cos 3x])`
= `(sin 9x + sin 7x - sin 9x - sin 3x)/(cos 3x + cos x - cos x + cos 7x)`
= `(sin 7x - sin 3x)/(cos 3x + cos 7x)`
= `(sin 7x - sin 3x)/(cos 7x + cos 3x)`
= `(2cos((7x + 3x)/2) * sin ((7x 3x)/2))/(2cos((7x + 3x)/2) * cos((7x - 3x)/2)`
= `(2 cos 5x * sin 2x)/(2 cos 5x * cos 2x)`
= `(sin 2x)/(cos 2x)`
= tan 2x
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