Calculus I - Trig Functions

Section 1.3 : Trig Functions

9. Determine the exact value of $\displaystyle \tan \left( {\frac{{15\pi }}{4}} \right)$ without using a calculator.

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First we can notice that $4\pi - \frac{\pi }{4} = \frac{{15\pi }}{4}$ and also note that $4\pi $ is two complete revolutions so the terminal line for $\frac{{15\pi }}{4}$ and $ - \frac{\pi }{4}$ represent the same angle. Also note that $ - \frac{\pi }{4}$ will form an angle of $\frac{\pi }{4}$ with the positive $x$-axis in the fourth 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 fourth quadrant is labeled with both $\frac{{15\pi }}{4}$ and $ - \frac{\pi }{4}$ and a dashed vertical line connects the point where it meets the circle to the point for $\frac{\pi }{4}$, showing the two points have the same $x$-coordinate and $y$-coordinates that differ only in sign.$

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The coordinates of the line representing $\frac{{15\pi }}{4}$ will be the same as the coordinates of the line representing $\frac{\pi }{4}$ except that the $y$ coordinate 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,

\[\tan \left( \frac{15\pi }{4} \right)=\frac{\sin \left( \frac{15\pi }{4} \right)}{\cos \left( \frac{15\pi }{4} \right)}=\frac{-{}^{\sqrt{2}}/{}_{2}}{{}^{\sqrt{2}}/{}_{2}}=-1\]