Question

Evaluate csc(13π).

Answer 1

first off, let's recall that a full-go-around on the circle is 2π.

so 13π is really 2π + 2π + 2π + 2π + 2π + 2π + π.

so it really goes around the circle 6 times, and then an extra +π, landing the angle at π, so csc(13π) is pretty much the same as csc(π).

$\bf csc(\pi )\implies \cfrac{1}{sin(\pi )}\implies \stackrel{\stackrel{und efined}{\downarrow }}{\cfrac{1}{0}}$