Question

If $f_n\left(\alpha \right)=\frac{\sin \alpha +\sin 3\alpha +\sin 5\alpha +.....+\sin (2n-1)\alpha }{\cos \alpha +\cos 3\alpha +\cos 5\alpha +.....+\cos (2n-1)\alpha }$ find $f_4\left(\pi /32\right)$.

Answer 1

Using,

$\sum _{k=0}^{n-1}\cos (a+kd)=\frac{\sin (n\times d/2)}{\sin \frac{d}{2}}\times \cos \left(\frac{2a+(n-1)d}{2}\right)$

& $\sum _{k=0}^{n-1}\sin (a+kd)=\frac{\sin \left(nd/2\right)}{\sin \frac{d}{2}}\times \sin \left(\frac{2a+(n-1)d}{2}\right)$

$f_n\left(\alpha \right)=\frac{\sin \alpha +\sin 3\alpha +.....+\sin (2n-1)\alpha }{\cos \alpha +\cos 3\alpha +....+\cos (2x-1)\alpha }$

Here, $a=\alpha$ & $d=2\alpha$

$\therefore f_n\left(\alpha \right)=\frac{\frac{\sin \left(nd/2\right)}{\sin \frac{d}{2}}\times \sin \left(\frac{2a+(n-1)d}{2}\right)}{\frac{\sin (n\times d/2)}{\sin \frac{d}{2}}\times \cos \left(\frac{2a+(n-1)d}{2}\right)}$

$=\frac{\sin (\alpha +(n-1)\alpha }{\cos (\alpha +(n-1)\alpha }$ [ As $\frac{\sin \left(nd/2\right)}{\sin \left(d/2\right)}\ne 0$

$=\tan \left(n\alpha \right)$

$\therefore f_4\left(32\right)=\tan \left(4.\frac{\pi }{32}\right)=\tan \frac{\pi }{8}=\sqrt{2}-1$