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Question
Prove that `sin theta/2 sin (7theta)/2 + sin (3theta)/2 sin (11theta)/2` = sin 2θ sin 5θ
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Solution
`sin theta/2 sin (7theta)/2 = 1/2[2sin theta/2 sin (7theta)/2]`
= `1/2[cos (7theta - theta)/2 - cos (7theta + theta)/2]`
= `1/2(cos 3theta - cos 4theta)`
`sin (3theta)/2 sin (11theta)/2 = /2[2sin (3theta)/2 sin (11theta)/2]`
= `1/2[cos (11theta - 3theta)/2 - cos (11theta + 3theta)/2]`
= `1/2[cos4theta - cos7theta]`
L.H.S = `sin theta/2 sin (7theta)/2 + sin (3theta)/2 sin (11theta)/2`
= `1/2[cos 3theta - cos 4theta) + 1/2[cos 4theta- cos 7theta]`
= `1/2[cos 3theta - cos 4theta + cos 4theta - cos 7theta]`
= ``1/2[cos3theta - cos 7theta]`
= `1/2[sin (7theta + 3theta)/2 sin (7theta - 3theta)/2]`
= sin 5θ sin 2θ
= R.H.S
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