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Question
Show that `((cos theta -cos 3theta)(sin 8theta + sin 2theta))/((sin 5theta - sin theta) (cos 4theta - cos 6theta))` = 1
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Solution
`((cos theta -cos 3theta)(sin 8theta + sin 2theta))/((sin 5theta - sin theta) (cos 4theta - cos 6theta)) = ((cos 3theta -cos theta)(sin 8theta + sin 2theta))/((sin 5theta - sin theta) (cos 6theta - cos 4theta))`
= `(2sin((3theta + theta)/2) sin ((theta - 3theta)/2) * 2sin((8theta + 20)/2) cos((8theta - 2theta)/2))/((2cos((5theta + theta)/2) * sin((5theta - theta)/2) * 2sin((6theta - 4theta)/2) sin((4theta - 6theta)/2)`
= `(sin 2theta* sin(- theta) *sin 5theta * cos 3theta)/(cos3theta * sin 2theta * sin 5theta * sin (- theta))`
= 1
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