Trig Functions with the Unit Circle Exercise 1

Answer: \(\tan\left(\dfrac{3\pi}{4}\right)=-1,\)  \(\tan(210^\circ) = -\dfrac{\sqrt{3}}{3},\) \(\tan(\pi) = 0\)

Solution Method: To find the exact value of tangent, we use the formula
    \begin{equation*}
        \tan(\theta)=\frac{\sin{\theta}}{\cos{\theta}} 
    \end{equation*}
So we can find the exact value of tangent by finding the exact values of sine and cosine. Using right triangles, we know that every \((x,y)\) point on the unit circle is equivalent to \((\cos{\theta},\sin{\theta})\). (see previous video) So to find the exact value of tangent at a special angle, we take the x and y coordinates from the unit circle and divide them. 

For \(\tan\left(\frac{3\pi}{4}\right),\) the coordinate at \(\frac{3\pi}{4}\) is \(\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right).\) So, 
    \begin{equation*} 
        \tan(\frac{3\pi}{4})= \frac{y}{x} = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = \left(\frac{\sqrt{2}}{2}\right) \left(-\frac{2}{\sqrt{2}}\right) = -1
    \end{equation*}

For \(\tan(210^\circ),\) the coordinate is \(\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\). So,
    \begin{equation*}
        \tan(210^\circ)= \frac{y}{x} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \left( \frac{1}{2} \right) \left( \frac{2}{\sqrt{3}} \right) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}
    \end{equation*}
We had to rationalize the fraction to get our final answer.

For \(\tan(\pi),\) \(\pi\) is a quadrantal angle located at (-1,0). So, 
    \begin{equation*}
        \tan(\pi) = \frac{y}{x} = \frac{0}{1} = 0
    \end{equation*}

Beware of Quadrantals: At the quadrantal angles \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\), the x-coordinate is 0, meaning in the tangent ratio, we are dividing by 0. Dividing by 0 means the fraction is undefined! So \(\tan\left(\frac{\pi}{2}\right)\) and \(\tan\left(\frac{3\pi}{2}\right)\) are undefined.