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Question
Prove that: `2 cos pi/13 cos (9pi)/13 + cos (3pi)/13 + cos (5pi)/13 = 0`
Solution
L.H.S. = `2 cos pi/13 cos (9pi)/13 + cos (3pi)/13 + cos (5pi)/13`
= `cos (pi/13 + (9pi)/13) + cos (pi/13 - (9pi)/13) + cos (3pi)/13 + cos (5pi)/13`
= `cos (10pi)/13 + cos (8pi)/13 + cos (3pi)/13 + cos (5pi)/13` [2cos x cos y = cos(x + y) + cos (x - y)]
= `cos (pi - (13pi)/13) + cos (pi - (8pi)/13) + cos (3pi)/13 + cos (5x)/13`
= `- cos (3pi)/13 - cos (5pi)/13 + cos (3pi)/13 + cos (5pi)/13 = 0` [∵ cos (π - θ) = -cosθ]
= 0 = R.H.S.
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