# How do you write csc(2x) / tanx in terms of sinx?

May 14, 2016

$\frac{1}{2 {\sin}^{2} \left(x\right)}$

#### Explanation:

Useful Trig ID's

Definitions of functions
$\csc \left(x\right) = \frac{1}{\sin} \left(x\right)$

$\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$

Sums of Angles Formula
$\sin \left(x + y\right) = \sin \left(x\right) \cos \left(y\right) + \cos \left(x\right) \sin \left(y\right)$
Which gives the double well known double angle formula
$\sin \left(2 x\right) = 2 \sin \left(x\right) \cos \left(x\right)$

We start with our ID, sub in the basic definition and use some fraction rules to get the following.

$\csc \frac{2 x}{\tan} \left(x\right) = \frac{\frac{1}{\sin} \left(2 x\right)}{\sin \frac{x}{\cos} \left(x\right)} = \frac{1}{\sin} \left(2 x\right) \cos \frac{x}{\sin} \left(x\right)$

We replace $\sin \left(2 x\right)$ with $2 \sin \left(x\right) \cos \left(x\right)$

$= \frac{1}{2 \sin \left(x\right) \cos \left(x\right)} \cos \frac{x}{\sin} \left(x\right)$

The cosine's cancel
$= \frac{1}{2 \sin \left(x\right)} \frac{1}{\sin} \left(x\right)$
leaving us with

$= \frac{1}{2 {\sin}^{2} \left(x\right)}$