A nice Trignometric Series

this is hard to write.(sry bcoz i dunno latex)

1) $log (1+x) = x - x^2+2 + x^3/3 ..........$

2)sin(rx) = e^(irx)-e^(-irx) /2i

3)sub .into eqn

4)therefore

2if(x) = (sigma)(1 to inf ) [ { (-1)^(r+1) (e^ix sinx)^r }/r }- {(-1)^(r+1) (e^(-ix) sinx)^r }/r}

from (1)

2if(x) = log(1+e^(ix)sinx ) - log(e^(-ix) sinx) = log[ (1+e^(ix)sinx ) /(1+e^(-ix)sinx ) ]

2if = log[ (cosxsinx +1 + isinxsinx )/(cosxsinx +1 - isinxsinx ) ]

2if= log[e^2iarg(cosxsinx +1 + isinxsinx ) ]....( bcoz it is (Ae^ik/Ae^-ik=e^i2k , k is arg)

hence

f=arg(cosxsinx +1 + isinxsinx )

f=arctan (1-cos2x / 2+sin2x) ..put double angle formula

put x=pi/12 and u get a=2,b=3,c=5 .....ANS=2+3+5

Note: I'm not very good at this problem, so my solution is somewhat long-winded.

$\quad \quad \;$ Hopefully there are no mistakes.

$f(x) = \sum\limits_{r=1}^\infty \frac{1}{r} (-1)^{r-1} \sin rx \sin^r x$

$f'(x) = \sum\limits_{r=1}^\infty \frac{1}{r} (-1)^{r-1} (r \cos rx \sin^r x + r \sin^{r-1}x \sin rx \cos x)$

$\quad \quad \; \; = \sum\limits_{r=1}^\infty (-\sin x)^{r-1} (\cos rx \sin x + \sin rx \cos x)$

$\quad \quad \; \; = \sum\limits_{r=1}^\infty (-\sin x)^{r-1} \sin (r+1)x$

Let $z = Cis x$ , then

$f'(x) = Im \sum\limits_{r=1}^\infty (-\sin x)^{r-1} z^{r+1}$

$\quad \quad \; \; = Im \sum\limits_{r=1}^\infty z^2 (-z \sin x)^{r-1}$

Since $|z\sin x|<1$ in this case,

$f'(x) = Im \frac {z^2}{1+z \sin x}$

$\quad \quad \; \; = Im \frac{\cos 2x + i\sin 2x}{ 1+\sin x \cos x +i\sin^2 x}$

$\quad \quad \; \; = Im \frac{(\cos 2x + i\sin 2x)}{1+\frac{1}{2} \sin 2x +i\sin^2 x} \frac{1+\frac{1}{2} \sin 2x - i\sin^2 x}{1+\frac{1}{2} \sin 2x - i\sin^2 x}$

$\quad \quad \; \; = \frac{-\cos 2x \sin^2 x + \sin 2x + \frac{1}{2} \sin^2 2x}{1 +\sin 2x + \frac{1}{4} \sin^2 2x + \sin^4 x}$

Note that $\sin^4 x = \sin^2 x(1-\cos^2 x) = \sin^2 x - \frac{1}{4} \sin^2 2x$

$\quad \quad \;$ and $-\cos 2x \sin^2 x = -(2 \cos^2 x -1)\sin^2 x = -\frac{1}{2} \sin^2 2x + \sin^2 x$

Hence,

$f'(x) = \frac{\sin 2x + \sin^2 x}{1+ \sin 2x + \sin^2 x} = 1 - \frac{1}{1+ \sin 2x + \sin^2 x}$

$f(\frac{\pi}{12})-f(0) = \int_{0}^{\frac{\pi}{12}} 1 - \frac{dx}{1+ \sin 2x + \sin^2 x}$

Since $f(0) = 0$ ,

$f(\frac{\pi}{12}) = \frac{\pi}{12} + \int_{0}^{\frac{\pi}{12}} \frac{-dx}{1+ \sin 2x + \sin^2 x}$

$\int_{0}^{\frac{\pi}{12}} \frac{-dx}{1+ \sin 2x + \sin^2 x} = \int_{0}^{\frac{\pi}{12}} \frac{-dx}{1+ 2\sin x \cos x + \sin^2 x}$

$\quad \quad \quad \quad \quad \quad \quad \; = \int_{0}^{\frac{\pi}{12}} \frac{-\csc^2 x \ dx}{\csc^2 x+ 2\cot x + 1}$

Let $u = \cot x + 1$ , then $du = -\csc^2 x \ dx$

$\qquad \qquad \qquad \qquad$ and $u^2 = \cot x + 2\cot x + 1 = \csc^2 x + 2\cot x$

The integral now becomes $\int_{\infty}^{3+\sqrt 3} \frac{du}{1+u^2} = \tan^{-1} [x]_{\infty}^{3+\sqrt 3}$

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \tan^{-1}(3+\sqrt 3) - \frac{\pi}{2}$

Note that $\tan^{-1}x + cot^{-1}x = \frac{\pi}{2}$ for $x>0$ ,

Hence the value of the definite integral is

$-\cot^{-1}(3+\sqrt 3) = -\tan^{-1} (\frac{1}{3+\sqrt 3}) = -\tan^{-1} (\frac{3- \sqrt 3}{6})$

Finally,

$f(\frac{\pi}{12}) = \frac{\pi}{12} -\tan^{-1} (\frac{3- \sqrt 3}{6})$

$\frac {a- \sqrt b}{c} = \tan( \frac{\pi}{12} -\tan^{-1} (\frac{3- \sqrt 3}{6}) )$

Solving, we obtain $\frac {a- \sqrt b}{c} = \frac {2- \sqrt 3}{5}$ ,

Answer $: 2+3+5 = \boxed{10}$