The Unit Circle

Other Trig Functions

We now want to combine what we just learned about using the unit circle to compute the values of \(\sin(\theta)\) and \(\cos(\theta)\) with the fundamental identities to compute the values of the other trigonometric functions. To do this, let's first revisit the illustration of a right triangle inside the unit circle. We saw earlier that we were able to define \(\sin(\theta)\) and \(\cos(\theta)\) in terms of the \(y\) and \(x\) coordinates, respectively. We can do the same thing with the other trig functions.

Here is a general process that we can follow to compute the value of any of the trigonometric functions at any of the "nice" angles or values. More specifically, this oultines a process to compute the trig values for angles corresponding to \(\theta\) = 0°, 30°, 45°, 60°, and 90° or to \(t = 0\), \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\), and \(\frac{\pi}{2}\).

1. Locate where the terminal side of the given angle is located on the unit circle. It may be helpful to sketch a quick graph. If the given angle is negative, you may want to apply an even-odd identity to rewrite the expression with a positive angle.
2. If the angle is greater than a full revolution, then determine the coterminal angle on the interval \([0, 2\pi)\), or between 0° and 360°.
3. If the angle is a quadrantal angle (coterminal to the \(x\) or \(y\) axis), then the terminal point is \((1,0)\), \((0,1)\), \((-1,0)\), or \((0,-1)\). However, if the terminal side is located within a quadrant (so it is not quadrantal), then determine its reference angle or reference number. The three "nice" coordinates in the 1st quadrant are \(\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\), \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\), and \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\).
4. If the given trig function is not sine or cosine, then use the appropriate reciprocal identity to rewrite it in terms of \(\sin(\theta)\) and/or \(\cos(\theta)\). Pick the \(x\) and/or \(y\) coordinate(s) that correspond to the quadrantal or reference angle and subsitute the value(s) into your expression. Remember \(x = \cos(\theta)\) and \(y = \sin(\theta)\).
5. Apply the sign based on the quadrant containing the terminal side. Simplify to get your answer.

This may sound like a lot, but don't let it intimidate you. You will get the hang of it. Let's work through some examples.

Example: Compute the exact values for (a) \(\csc(30^{\circ})\), (b) \(\sec\left(-\frac{\pi}{4}\right)\), and (c) \(\cot\left(\frac{10\pi}{3}\right)\).

Solution

a) Compute the exact value of \(\csc(30^{\circ})\).

A 30° angle is already an acute angle in the 1st quadrant, and its corresponding terminal point is \(\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\).

Next, we want to use a reciprocal identity to rewrite \(\csc(\theta)\) in terms of \(\sin(\theta)\). More specifically, we have \(\csc(\theta) = \frac{1}{\sin(\theta)}\). Since \(\sin(\theta)\) is the \(y\)-coordinate, we have \(\sin(30^{\circ}) = \frac{1}{2}\). Lastly, we know \(\sin(\theta) > 0\) in the 1st quadrant. So let's put this all together.

\[\begin{align*} \csc(30^{\circ}) &= \frac{1}{\sin(30^{\circ})} \\ &= \frac{1}{\frac{1}{2}} \\ &= 1 \cdot \frac{2}{1} \\ &= 2 \end{align*}\]

So, we get \(\csc(30^{\circ}) = 2\).

b) Compute the exact value of \(\sec\left(-\frac{\pi}{4}\right)\).

A angle of \(-\frac{\pi}{4}\) is a clockwise rotation into the 4th quadrant. It has a \(\frac{\pi}{4}\) reference angle with corresponding terminal point \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\).

Since \(\sec(\theta)\) is the reciprocal of \(\cos(\theta)\), we can use the reciprocal identity \(\sec(\theta) = \frac{1}{\cos(\theta)}\) and the even-odd identity \(\cos(-\theta) = \cos(\theta)\). Since \(\cos(\theta)\) is the \(x\)-coordinate, we have \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\). We also know that \(\cos(\theta) > 0\) in the 4th quadrant.

\[\begin{align*} \sec\left(-\frac{\pi}{4}\right) &= \frac{1}{\cos\left(-\frac{\pi}{4}\right)} \\ &= \frac{1}{\cos\left(\frac{\pi}{4}\right)} \\ &= \frac{1}{\frac{\sqrt{2}}{2}} \\ &= 1 \cdot \frac{2}{\sqrt{2}} \\ &= \frac{2}{\sqrt{2}} \left(\frac{\sqrt{2}}{\sqrt{2}}\right) \\ &= \frac{2\sqrt{2}}{2} \\ &= \sqrt{2} \end{align*}\]

So, we get \(\sec\left(-\frac{\pi}{4}\right) = \sqrt{2}\).

c) Compute the exact value of \(\cot\left(\frac{10\pi}{3}\right)\).

A full revolution is \(\frac{6\pi}{3}\) because \(\frac{6\pi}{3} = 2\pi\). This means that \(\frac{10\pi}{3}\) is coterminal to \(\frac{4\pi}{3}\), which is located in the 3rd quadrant. It has a reference angle of \(\frac{\pi}{3}\) which has a terminal point of \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\).

We can use the reciprocal identity \(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\) to rewrite \(\cot(\theta)\) in terms of \(\cos(\theta)\) and \(\sin(\theta)\). We will need to use both the \(x\) and \(y\) coordinates of the terminal point since we are using both \(\cos(\theta)\) and \(\sin(\theta)\). We also know that \(\cot(\theta) > 0\) in the 3rd quadrant even though both \(\cos(\theta)\) and \(\sin(\theta)\) are negative.

\[\begin{align*} \cot\left(\frac{10\pi}{3}\right) &= \frac{\cos\left(\frac{10\pi}{3}\right)}{\sin\left(\frac{10\pi}{3}\right)} \\ &= \frac{-\cos\left(\frac{\pi}{3}\right)}{-\sin\left(\frac{\pi}{3}\right)} \\ &= \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} \\ &= -\frac{1}{\cancel{2}} \cdot -\frac{\cancel{2}}{\sqrt{3}} \\ &= \frac{1}{\sqrt{3}} \end{align*}\]

So, we get \(\cot\left(\frac{10\pi}{3}\right) = \frac{1}{\sqrt{3}}\).