Question

Why is tan(theta) equal to tan(theta + 2pi)

Answer 1

Recall that tangent is the ratio of sine over cosine

tan = sin/cos

This means we'll have sin(theta+2pi) up top and cos(theta+2pi) in the bottom

$\tan(\theta+2\pi) = \frac{\sin(\theta+2\pi)}{\cos(\theta+2\pi)}$

Now because both sine and cosine have a period of 2pi, this means,

$\sin(\theta+2\pi) = \sin(\theta)\\\cos(\theta+2\pi) = \cos(\theta)$

The graph of each repeats itself every 2pi units, which is why we're back to the original version of each.

So,

$\tan(\theta+2\pi) = \frac{\sin(\theta+2\pi)}{\cos(\theta+2\pi)}\\\\\tan(\theta+2\pi) = \frac{\sin(\theta)}{\cos(\theta)}\\\\\tan(\theta+2\pi) = \tan(\theta)$

This seems to suggest that tangent also has a period of 2pi. This is false or misleading. It turns out the period of tangent is pi. The proof of this is a bit more involved. See the screenshot below to see those steps.