The unit circle is defined as $x^2 + y^2 = 1$, which is a circle with radius $1$, centered at the origin $(0, 0)$. It is used in trigonometry to simplify finding values of trig functions. An empty unit circle is below with markings to be filled in:

Blank_Unit_Circle.PDF

Practice Problems

1. Find the radian value of $120^\circ$.
2. Find the coordinates corresponding to the radian value $\dfrac{7\pi}6$.
3. Find all angles $\theta$ between $0$ and $2\pi$ such that $|\sin \theta| = \dfrac{\sqrt{3}}2$.
4. In which quadrants is $\tan \theta$ positive?

Solutions

1. $$120^\circ\cdot\frac{\pi}{180^\circ} = \boxed{\frac{2\pi}{3}}$$
2. We wish to find the $x$-coordinate $\cos\dfrac{7\pi}{6}$ and the $y$-coordinate $\sin\dfrac{7\pi}{6}$. First note that
   $$\pi< \dfrac{7\pi}{6}<\dfrac{3\pi}{2}$$
   so the angle lies in quadrant III. In quadrant III both $x$- and $y$-coordinates are negative. The reference angle of $\dfrac{7\pi}{6}$ is
   $$\dfrac{7\pi}{6}-\pi = \dfrac{\pi}{6}$$
   Therefore, the coordinates are
   $$ \bigg(\cos\frac{7\pi}{6},\sin\frac{7\pi}{6}\bigg) = \bigg(-\cos\frac{\pi}{6},-\sin\frac{\pi}{6}\bigg) = \boxed{\bigg(-\frac{\sqrt3}{2},\frac12\bigg)} $$
3. If
   $$|\sin\theta| = \dfrac{\sqrt3}{2}$$
   then if $\theta’$ is the reference angle, then
   $$\sin\theta’ =\frac{\sqrt3}{2}$$
   Since $\theta’$ is less than $\dfrac{\pi}{2}$, we can use the $30-60-90$ right triangle to see that
   $$\theta’ = \frac{\pi}{3}$$
   The value of $\theta$ in quadrant II is
   $$\pi – \theta’ = \pi-\frac{\pi}{3} = \frac{2\pi}{3}$$
   The value of $\theta$ in quadrant III is
   $$\pi + \theta’ = \pi+\frac{\pi}{3} = \frac{4\pi}{3}$$
   The value of $\theta$ in quadrant IV is
   $$2\pi – \theta’ = 2\pi-\frac{\pi}{3} = \frac{5\pi}{3}$$
   Thus, all the angles are
   $$ \boxed{\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}} $$
4. Since
   $$ \tan\theta = \frac{\sin\theta}{\cos\theta} $$
   $\tan\theta$ is positive only if both $\sin\theta$ and $\cos\theta$ are positive or both $\sin\theta$ and $\cos\theta$ are negative. $\sin\theta$ and $\cos\theta$ represent $y$- and $x$-coordinates. Both $x$- and $y$-coordinates are positive in quadrant I and both are negative in quadrant III. Therefore, $\tan\theta$ is positive in quadrants I and III. Note that $\tan\theta$ is 0 if $\sin\theta = 0$, which happens on the $x$-axis. In addition, $\tan\theta$ is undefined if $\cos\theta = 0$. which happens on the $y$-axis.