Question

Use reference angles to evaluate sec(13Pi/4) enter exact answers

Answer 1

Answer:

-√2

Explanation:

Given the trigonometric function:

$\sec \mleft(\frac{13\pi}{4}\mright)$

• The function cos is periodical with a period of 2π

,

• The reference angle is an angle less than 2π and that is positive.

We can write this as:

$\sec \mleft(\frac{13\pi}{4}-2\pi\mright)=\sec \mleft(\frac{5\pi}{4}\mright)$

Since the angle π is in the third quadrant, subtract π from it.

$\begin{gathered} \sec \mleft(\frac{5\pi}{4}\mright)=\sec \mleft(\frac{5\pi}{4}-\pi\mright) \\ =\sec \mleft(\frac{\pi}{4}\mright) \end{gathered}$

• Secant is the inverse of cosine.

,

• Note that cosine is negative in Quadrant III.

Thus:

$\begin{gathered} \sec \mleft(\frac{\pi}{4}\mright)=\frac{1}{-\cos \mleft(\frac{\pi}{4}\mright)} \\ =1\div-\frac{1}{\sqrt[]{2}} \\ =1\times-\sqrt[]{2} \\ =-\sqrt[]{2} \end{gathered}$

Therefore:

$\sec (\frac{13\pi}{4})=-\sqrt[]{2}$