What's Your Approach

A faster approach is to recognize that we have a telescoping sum. Because $\frac{1}{2} \tan \frac{x}{2} = \frac{1}{2} \cot \frac{x}{2} - \cot x$ , we can see that the partial sums telescope to

$- \cot x + \frac{1}{ 2^n} \cot \frac{ x}{ 2^n }.$

Then, taking limits as $n \rightarrow \infty$ , we obtain $- \cot x + \frac{ 1}{ x}$ .

We can thus evaluate that

$f ( \frac{ 3 \pi } { 8} ) = \frac{ 8}{ 3 \pi } + 1 - \sqrt{ 2} .$

This "explains" why the formula looks so strange, involving both a trigonometric value and an algebraic value.

$P(n) = cos\dfrac{x}{2^{n}}cos\dfrac{x}{2^{n-1}}.... cos\dfrac{x}{2}$

$lim_{n \to \infty}P(x)$

$P(n) = \dfrac{2sin\dfrac{x}{2^{n}}cos\dfrac{x}{2^{n}}cos\dfrac{x}{2^{n-1}}..... cos\dfrac{x}{2}}{2sin\dfrac{x}{2^{n}}}$

$2sinxcosx=sin2x$

thus we get(multiplying and divivding by 2 n times and applying the above identity in each step)

$lim_{n \to \infty}P(n) = lim_{n \to \infty}\dfrac{sinx}{2^{n}sin\dfrac{x}{2^{n}}}$

$lim_{n \to \infty}P(n) = lim_{n \to \infty}\dfrac{sinx}{2^{n}\dfrac{sin\dfrac{x}{2^{n}}}{\dfrac{x}{2^{n}}}\times\dfrac{x}{2^{n}}}$

$lim_{n \to \infty}P(n) = \dfrac{sinx}{x}$

Thus,

$cos\dfrac{x}{2}cos\dfrac{x}{2^{2}}cos\dfrac{x}{2^{3}}..... = \dfrac{sinx}{x}$

$logcos\dfrac{x}{2} + logcos\dfrac{x}{2^{2}}+ logcos\dfrac{x}{2^{3}}+.... = logsinx - logx$

Differentiating,

$- (\dfrac{1}{2}tan\dfrac{1}{2} + \dfrac{1}{4}tan\dfrac{1}{4} + \dfrac{1}{8}tan\dfrac{1}{8}+......) = cotx - \dfrac{1}{x}$

$\dfrac{1}{2}tan\dfrac{1}{2}+ \dfrac{1}{4}tan\dfrac{1}{4} + \dfrac{1}{8}tan\dfrac{1}{8} + ............ = - cotx + \dfrac{1}{x}$

$f(x) = \dfrac{1}{x} - cotx$

$f(\dfrac{3\pi}{8}) = \dfrac{8}{3\pi} - cot(\dfrac{\pi}{2} - \dfrac{\pi}{8})$

$f(\dfrac{3\pi}{8}) = \dfrac{8}{3\pi} - tan\dfrac{\pi}{8}$

$tan2x = \dfrac{2tanx}{1 - tan^{2}x}$

$tan\dfrac{\pi}{4} = \dfrac{2tan\dfrac{\pi}{8}}{ 1- tan^{2}\dfrac{\pi}{8}}$

$1 - tan^{2}\dfrac{\pi}{8} = 2tan\dfrac{\pi}{8}$

$2 = tan^{2}\dfrac{\pi}{8} + 2tan\dfrac{\pi}{8} + 1$

$1 + tan\dfrac{\pi}{8} = \pm \sqrt{2}$

$tan\dfrac{\pi}{8} = \sqrt{2} - 1$ ( as in first quadrant)

$f(\dfrac{3\pi}{8}) = \dfrac{8}{3\pi} - ( \sqrt{2} - 1)$

$f(\dfrac{3\pi}{8}) = \dfrac{8}{3\pi} + 1 - \sqrt{2}$

Here is similar but shorter method I want to share

$\displaystyle f(x) = \frac{1}{2} tan(\frac{x}{2}) + \frac{1}{4} tan(\frac{x}{4}) \ldots$

Integrating

$\displaystyle \int f(x) = - ln(cos(\frac{x}{2})) - ln(cos(\frac{x}{4})) - \ldots$

$\displaystyle \int f(x) = - \left[ ln \left( cos(\frac{x}{2}). cos(\frac{x}{4}).cos(\frac{x}{8}) \ldots \right )\right]$

We have an identity

$\displaystyle cos(y).cos(2y).cos(4y)\ldots cos(2^{n-1}y) = \dfrac{sin(2^{n}y)}{2^{n} sin(y)}$

Here $y = \frac{x}{2^{n}}$

We get

$\displaystyle \int f(x) = - ln(\frac{sin(x)}{2^{n} sin(\frac{x}{2^{n}})})$

Here $n \to \infty$

$\displaystyle \int f(x) = -ln(\frac{sinx}{x})$

Differentiating we finally get

$\displaystyle f(x) = \frac{1}{x} - cot(x)$

$f(\frac{3\pi}{8}) = \boxed{\boxed{ \frac{8}{3\pi} + 1 - \sqrt{2} }}$

Hey I got the negative of the answer, and then just assumed it was the negative of the correct answer and got 15. Like i got f(x)=cotx-1/x. How does that extra minus sign pop up?