Question

what is $\sec(\frac{13\pi}{4})$?

$-\sqrt{?}$

Explanation:

Step1: Use the periodicity of the secant function

The period of \(y = \sec(x)\) is \(2\pi\). So, \(\sec(\frac{13\pi}{4})=\sec(\frac{13\pi}{4}- 3\pi)=\sec(\frac{13\pi - 12\pi}{4})=\sec(\frac{\pi}{4})\).

Step2: Use the identity \(\sec(x)=\frac{1}{\cos(x)}\)

We know that \(\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\), so \(\sec(\frac{\pi}{4})=\frac{1}{\cos(\frac{\pi}{4})}\).
Substitute \(\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\) into the formula: \(\sec(\frac{\pi}{4})=\frac{1}{\frac{\sqrt{2}}{2}}=\sqrt{2}\).
But \(\frac{13\pi}{4}\) is in the third - quadrant. In the third - quadrant, \(\cos(x)<0\) and \(\sec(x)=\frac{1}{\cos(x)}<0\). So \(\sec(\frac{13\pi}{4})=-\sqrt{2}\)

Answer:

\(2\)