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प्रश्न
Prove that:
2 cos `pi/13` cos \[\frac{9\pi}{13} + \text{cos} \frac{3\pi}{13} + \text{cos} \frac{5\pi}{13}\] = 0
उत्तर
LHS = 2 cos `pi/13` cos \[\frac{9\pi}{13} + \text{cos} \frac{3\pi}{13} + \text{cos} \frac{5\pi}{13}\]
`= 2 cos pi/13 cos (9pi)/13 + 2(cos ((3pi)/13 + (5pi)/13)/2) xx (cos ((3pi)/13 - (5pi)/13)/2)`
`[∵ cos "C" + cos "D" = 2 cos (("C + D")/2) cos (("C - D")/2)]`
`= 2 cos pi/13 cos (9pi)/13 + 2(cos ((8pi)/13)/2) xx (cos (-(2pi)/13)/2)`
`= 2 cos pi/13 cos (9pi)/13 + 2cos (4pi)/13 cos ((-pi)/13)`
[∵ cos(-θ) = cos θ]
`= 2 cos pi/13 cos (9pi)/13 + 2cos (4pi)/13 cos pi/13`
`= 2 cos pi/13 (cos (9pi)/13 + cos (4pi)/13)`
[take 2 cos `pi/3` as common]
`= 2 cos pi/13 (2 cos (((9pi + 4pi)/13))/2 cos ((9pi - 4pi)/13)/2)`
`= 2 cos pi/13 (2 cos (13pi)/(13 xx 2) cos (5pi)/(13 xx 2))`
`= 2 cos pi/13 (2 cos pi/2 cos (5pi)/2)`
`= 2 cos pi/13 (0 xx cos (5pi)/(13 xx 2))`
= 0 = RHS
Hence proved.
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