Advertisements
Advertisements
Question
Prove that:
2 cos `pi/13` cos \[\frac{9\pi}{13} + \text{cos} \frac{3\pi}{13} + \text{cos} \frac{5\pi}{13}\] = 0
Advertisements
Solution
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.
APPEARS IN
RELATED QUESTIONS
Prove that:
Show that :
Show that:
sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0
Express each of the following as the product of sines and cosines:
cos 12x + cos 8x
Prove that:
cos 20° cos 100° + cos 100° cos 140° − 140° cos 200° = −\[\frac{3}{4}\]
Prove that:
Prove that:
Prove that:
If sin (B + C − A), sin (C + A − B), sin (A + B − C) are in A.P., then cot A, cot B and cot Care in
If sin x + sin y = \[\sqrt{3}\] (cos y − cos x), then sin 3x + sin 3y =
