Calculus I - Trig Functions

Section 1.3 : Trig Functions

11. Determine the exact value of $\displaystyle \sec \left( {\frac{{29\pi }}{4}} \right)$ without using a calculator.

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First we can notice that $\frac{{5\pi }}{4} + 6\pi = \frac{{29\pi }}{4}$ and recalling that $6\pi $ is three complete revolutions we can see that $\frac{{29\pi }}{4}$ and $\frac{{5\pi }}{4}$ represent the same angle. Next, note that $\pi + \frac{\pi }{4} = \frac{{5\pi }}{4}$ and so the line representing $\frac{{5\pi }}{4}$ will form an angle of $\frac{\pi }{4}$ with the negative $x$-axis in the third quadrant and we’ll have the following unit circle for this problem.

$A unit circle with the angles $\frac{\pi }{6}$, $\frac{\pi }{4}$ and $\frac{\pi }{3}$ drawn in the first quadrant and labeled with the coordinates $\left( {\frac{{\sqrt 3 }}{2},\frac{1}{2}} \right)$, $\left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right)$ and $\left( {\frac{1}{2},\frac{{\sqrt 3 }}{2}} \right)$ where they meet the circle. The axes are labeled with the angles 0 and $2\pi $, $\frac{\pi }{2}$, $\pi $ and $\frac{{3\pi }}{2}$ and the points (1,0), (0,1), (-1,0) and (0,-1). A line in the third quadrant is labeled with both $\frac{{29\pi }}{4}$ and $\frac{{5\pi }}{4}$ and lies on the same line through the origin as the angle $\frac{\pi }{4}$, so the coordinates of the two points on the circle differ only in sign.$

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The line representing $\frac{{95\pi }}{4}$ is a mirror image of the line representing $\frac{\pi }{4}$ and so the coordinates for $\frac{{29\pi }}{4}$ will be the same as the coordinates for $\frac{\pi }{4}$ except that both coordinates will now be negative. So, our new coordinates will then be $\left( { - \frac{{\sqrt 2 }}{2}, - \frac{{\sqrt 2 }}{2}} \right)$ and so the answer is,

\[\sec \left( \frac{29\pi }{4} \right)=\frac{1}{\cos \left( \frac{29\pi }{4} \right)}=\frac{1}{-{}^{\sqrt{2}}/{}_{2}}=-\frac{2}{\sqrt{2}}=-\sqrt{2}\]