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) = (\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) = (\sin(\theta+2\pi))/(\cos(\theta+2\pi))\\\\\tan(\theta+2\pi) = (\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.