Question

If the value of
(sinx+sin3x+sin5x+.....+sin17x)/(cosx+cos3x+cos5x+.....+cos17x)
when x=pi/24 is of the from sqrt(a) + b, where a,b belong to integers , then a+b= ?

Answer 1

sinx + sin17x = 2sin9xcos8x
sin3x + sin15x = 2sin9xcos6x
sin5x + sin13x = 2sin9xcos4x
sin7x + sin11x = 2sin9xcos2x
sin9x

--> Numerator = sin9x (2cos8x + 2cos6x + 2cos4x + 2cos2x + 1)

cosx + cos17x = 2cos9xcos8x
cos3x + cos15x = 2cos9xcos6x
cos5x + cos13x = 2cos9xcos4x
cos7x + cos11x = 2cos9xcos2x
cos9x

--> Denumerator = cos9x (2cos8x + 2cos6x + 2cos4x + 2cos2x + 1)

So we have\[\frac{ Num }{ Den } = \frac{\sin9x}{\cos9x} = \tan(9x) \]

Then x = pi/24 so we have\[\tan(9\pi/24) = \tan (3\pi/8) \approx 2.414 = \sqrt a + b\]

Until now it is perfect except we cannot solve for a + b =...? with a,b are integers.

I think the question should be in form of \[\sqrt {a+b}\]

Answer 2

yes you started off gud and i got dat ..! but it was a integer type question so even dont have options or choices for that ques.

Answer 3

Approximation solution might be a better answer. :)

I think b = 1, a = 2

Then we have \[\sqrt{a} + b = \sqrt{2} + 1 \approx 1.414 + 1 = 2.414 = \tan(3\pi/8)\]

Haha, I think that is the only good choice of answer

So a + b = 3 with integer a = 2, and b = 1 ^_^ BRAVO

Answer 4

there you go ...bravo ..! thank you very much ..!