# Thread: Help with the first three terms of this infinite series please

1. ## Help with the first three terms of this infinite series please

Hi,

I'd really appreciate some help figuring out the terms of this infinite sum:

$\sum_{n=0}^{\infty }c_n$

where

$c_0=k=\sin\theta$

$k'\equiv \cos\theta$

$k_{n+1}=\frac{1-k'_n}{1+k'_n}$

Would the first three terms be...

$\sin\theta+\frac{1-\cos(\arcsin\theta)}{1+\cos(\arcsin\theta)}+\frac{ \left (1-\cos\left ( \arcsin\left ( \frac{1-\cos(\arcsin\theta)}{1+\cos(\arcsin\theta)} \right )\right ) \right )}{\left (1+\cos\left ( \arcsin\left ( \frac{1-\cos(\arcsin\theta)}{1+\cos(\arcsin\theta)} \right )\right ) \right )}$

?

Thanks

2. I don't see why you have arcsin in your answer. You have:

$
k_0 = sin \theta
$

$
k_0' = cos \theta
$

$
k_1 = \frac {1- k_0'} {1+ k_o'} = \frac {1+cos \theta} {1- cos\theta}
$

$
k'_1 = \frac {2 sin \theta} {(1-cos \theta)^2}
$

$
k_2 = \frac {1+ \frac {2 sin \theta} {(1-cos \theta)^2}} {1 - \frac {2 sin \theta} {(1-cos \theta)^2}} = \frac {1 - 2 cos \theta + 2 sin \theta + cos^2 \theta} {1 - 2 cos\theta -2 sin \theta + cos^2 \theta}
$