Calculus I - Trig Functions

Section 1.3 : Trig Functions

6. Determine the exact value of $\displaystyle \sec \left( { - \frac{{11\pi }}{6}} \right)$ without using a calculator.

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First, we can notice that $\frac{\pi }{6} - 2\pi = - \frac{{11\pi }}{6}$ and so (remembering that negative angles are rotated clockwise) we can see that the terminal line for $ - \frac{{11\pi }}{6}$ will form an angle of $\frac{\pi }{6}$ with the positive $x$-axis in the first quadrant. In other words, $ - \frac{{11\pi }}{6}$ and $\frac{\pi }{6}$ represent the same angle. So, 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). The line for the angle $\frac{\pi }{6}$ is also labeled $ - \frac{{11\pi }}{6}$, showing that the two angles end on the same line.$

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Because the two angles $ - \frac{{11\pi }}{6}$ and $\frac{\pi }{6}$ have the same coordinates the answer is,

\[\sec \left( -\frac{11\pi }{6} \right)=\frac{1}{\cos \left( -\frac{11\pi }{6} \right)}=\frac{1}{{}^{\sqrt{3}}/{}_{2}}=\frac{2}{\sqrt{3}}\]