Question

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

Answer 1

-√2

Step-by-step explanation:

Given the trigonometric function:

`\sec \mleft((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((13\pi)/(4)-2\pi\mright)=\sec \mleft((5\pi)/(4)\mright)`

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

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

• Secant is the inverse of cosine.

,

• Note that cosine is negative in Quadrant III.

Thus:

`\begin{gathered} \sec \mleft((\pi)/(4)\mright)=(1)/(-\cos \mleft((\pi)/(4)\mright)) \\ =1/-\frac{1}{\sqrt[]{2}} \\ =1*-\sqrt[]{2} \\ =-\sqrt[]{2} \end{gathered}`

Therefore:

`\sec ((13\pi)/(4))=-\sqrt[]{2}`