# How do you find the asymptote for y = csc (4x + pi)?

Feb 12, 2018

See below.

#### Explanation:

Identity:

$\textcolor{red}{\boldsymbol{\csc x = \frac{1}{\sin} x}}$

Vertical asymptotes occur where $\frac{1}{\sin} x$ is undefined.

i.e. where $\sin x = 0$

So:

1/(sin(4x+pi) will be undefined when. $\sin \left(4 x + \pi\right) = 0$

$\therefore$

4x+pi=arcsin(sin(4x+pi)=arcsin(0)=0,pi,2pi etc.

Using $2 \pi$

$4 x + \pi = 2 \pi \implies x = \frac{\pi}{4}$

We can write all solutions as:

$n \frac{\pi}{4}$

Where n is an integer.

This is where all vertical asymptotes occur:

This is confirmed by the graph of $\csc \left(4 x + \pi\right)$