Calculus I - Trig Functions

Section 1.3 : Trig Functions

1. Determine the exact value of $\displaystyle \cos \left( {\frac{{5\pi }}{6}} \right)$ without using a calculator.

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First, we can notice that $\pi - \frac{\pi }{6} = \frac{{5\pi }}{6}$ and so the terminal line for $\frac{{5\pi }}{6}$ will form an angle of $\frac{\pi }{6}$ with the negative $x$-axis in the second 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). The angle $\frac{{5\pi }}{6}$ is drawn in the second quadrant and a dashed horizontal line connects the point where it meets the circle to the point for $\frac{\pi }{6}$, showing the two points have the same $y$-coordinate and $x$-coordinates that differ only in sign.$

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The coordinates of the line representing $\frac{{5\pi }}{6}$ will be the same as the coordinates of the line representing $\frac{\pi }{6}$ except that the $x$ coordinate will now be negative. So, our new coordinates will then be $\left( { - \frac{{\sqrt 3 }}{2},\frac{1}{2}} \right)$ and so the answer is,

\[\cos \left( {\frac{{5\pi }}{6}} \right) = - \frac{{\sqrt 3 }}{2}\]