# How do you find the asymptotes for  f(x)=tan2x?

##### 1 Answer
Dec 19, 2016

Recall that $y = \tan x$ can be written as $y = \sin \frac{x}{\cos} x$. Then there will be asymptotes whenever $\cos x = 0$, since we cannot have the denominator equal $0$ without making the function undefined in the real number system.

Similarly, $f \left(x\right) = \tan 2 x$ can be rewritten as $f \left(x\right) = \frac{\sin 2 x}{\cos 2 x}$. We need to find the values of $x$ that make $\cos 2 x = 0$.

$\cos 2 x = 0$

$2 x = \arccos \left(0\right)$

$2 x = \frac{\pi}{2} , \frac{3 \pi}{2}$

$x = \frac{\pi}{4} \mathmr{and} \frac{3 \pi}{4}$

Don't forget to add the periodicity. The function $y = \cos 2 x$ has a period of $\pi$, so there will be asymptotes in $\tan \left(2 x\right)$ whenever $x = \frac{\pi}{4} + \pi n$ or $x = \frac{3 \pi}{4} + \pi n$, where $n$ is an integer.

Hopefully this helps!