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Question
Prove that cosθ `cos theta/2 - cos 3theta cos (9theta)/2` = sin 7θ sin 8θ.
[Hint: Express L.H.S. = `1/2[2costheta cos theta/2 - 2 cos 3theta cos (9theta)/2]`
Solution
L.H.S. cosθ `cos theta/2 - cos 3theta cos (9theta)/2`
= `1/2[2 cos theta cos theta/2] - 1/2[2 cos 3theta cos (9theta)/2]`
= `1/2[cos(theta + theta/2) + cos(theta - theta/2)] - 1/2[cos(3theta + (9theta)/2) + cos(3theta - (9theta)/2)]`
= `1/2[cos (3theta)/2 + cos theta/2 - cos (15theta)/2 - cos((-3theta)/2)]`
= `1/2[cos (3theta)/2 + cos theta/2 - cos (15theta)/2 - cos (3theta)/2]` ......[∵ cos (-θ) = cos θ]
= `1/2[cos theta/2 - cos (15theta)/2]`
= `1/2[-2sin(theta/2 + (15theta)/2).sin(theta/2 - (15theta)/2)]`
= -sin 8θ sin (-7θ)
= sin 7θ sin 8θ .......[∵ sin (-θ) = -sin θ]
L.H.S. = R.H.S.
Hence proved.
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