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Sagot :
Answer:
-√2
Explanation:
Given the trigonometric function:
[tex]\sec \mleft(\frac{13\pi}{4}\mright)[/tex]• 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:
[tex]\sec \mleft(\frac{13\pi}{4}-2\pi\mright)=\sec \mleft(\frac{5\pi}{4}\mright)[/tex]Since the angle π is in the third quadrant, subtract π from it.
[tex]\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}[/tex]• Secant is the inverse of cosine.
,• Note that cosine is negative in Quadrant III.
Thus:
[tex]\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}[/tex]Therefore:
[tex]\sec (\frac{13\pi}{4})=-\sqrt[]{2}[/tex]
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