Can a Suitable Small Angle Formula Solve This Summation Problem?

[ghostyc]

Homework Statement

S_N(x)= \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\sin ((2 n-1)x)}{2 n-1}

By considering a suitable small angle formula show that the value of the sum at this point is

S_N \Big( \frac{\pi}{2 N} \Big)=\frac{2}{\pi} \int_0^{\pi} \frac{\sin (\mu)}{\mu} \; d{\mu}

Homework Equations

i have no idea how to get the suitable small angle formula working with this problem

The Attempt at a Solution

I have shown that

S_N(x)

can be written as

S_N(x)=\frac{2}{\pi} \int_0^{x} \frac{\sin (2 N t)}{\sin (t) } \; d{t}

my guess for suitable small angle formula is

\sin (x) \approx x when x is small


Thank you for any help

 

[ghostyc]

anyone got any ideas? :P