Section 2.2

Definition 6

A reference angle for \(\theta\), is denoted \(\theta'\), is the acute angle made by the terminal side of the angle \(\theta\) and the \(x\)-axis.

Example 2

Find the reference angle for the following angle measures.

If \(\theta = 170^{\circ}\), find \(\theta'\).

Solution:

\[\theta' = 180-170=10\]

Therefore, \(\theta'=10^{\circ}\).

If \(\theta = 200^{\circ}\), find \(\theta'\).

Solution:

\[\theta'=200-180 = 20\]

Therefore, \(\theta'=20^{\circ}\)

If \(\theta = 400^{\circ}\), find \(\theta'\).

Solution:

First, we must find the coterminal angle:

\[\theta_c=400-360=40\]

Next, we find the reference angle for \(40^{\circ}\). The reference angle for \(40^{\circ}\) is \(40^{\circ}\). Therefore, \(\theta' = 40^{\circ}\).

If \(\theta = 830^{\circ}\), find \(\theta'\)

Solution:

First, we need to find the coterminal angle:

\[\begin{align*} 830-360 & = 470\\ 470-360 & = 110\\ \theta_c & = 110^{\circ} \end{align*}\]

Next, we find the reference angle for \(110^{\circ}\).

\[180-110=70\]

Therefore, \(\theta' = 70^{\circ}\).

If \(\theta = -120^{\circ}\), find \(\theta'\).

Solution:

First, we must find the coterminal angle:

\[-120+360 = 240\]

Therefore, \(\theta_c=240^{\circ}\).

Next, we will find the reference angle for \(240^{\circ}\).

\[360-240 = 60\]

Therefore, \(\theta' = 60^{\circ}\).

If \(\theta = -750^{\circ}\), find \(\theta'\).

Solution:

First, we will find the coterminal angle:

\[\begin{align*} -750+360 & = -390\\ -390+360 & = -30\\ -30 +360 & = 330\\ \theta_c & = 330^{\circ} \end{align*}\]

Next, we will find the reference angle for \(330^{\circ}\).

\[360-330 = 30\]

Therefore, \(\theta'=30^{\circ}\).

Recall the table of values for cosine and sine from the previous section and attempt the following example.

Example 3

Find the exact values of the following:

\(\cos(135^{\circ})\)

Solution:

If \(\theta = 135^{\circ}\), then

\[\theta' = 180-135 = 45^{\circ}\]

Therefore, the size of \(\cos(135^{\circ})\) is \(\frac{\sqrt{2}}{2}\). That is, \(|\cos(135^{\circ})|=\frac{\sqrt2}{2}\).

Since the terminal side of \(135^{\circ}\) is in the second quadrant and cosine is negative in the second quadrant we finally know

\[\cos(135^{\circ}) = - \cos(45^{\circ}) = -\frac{\sqrt2}{2}\]

\(\sin(495^{\circ})\)

{dropdown} Solution:

First, we will find the coterminal angle:

\[495 - 360 = 135\]

Which means \(\theta_c = 135^{\circ}\) and from previous example we then know \(\theta'=45^{\circ}\). Next, since sine is positive in the second quadrant we know that the sign of the value will be positive. Therefore,

\[\sin(495^{\circ})=\sin(45^{\circ}) = \frac{\sqrt2}{2}\]

\(\cos(240^{\circ})\)

Solution:

First, we will find the reference angle

\[240-180 = 60\]

Therefore, \(\theta'=60\).

Since the terminal side of \(240^{\circ}\) is in quadrant three and cosine is negative in quadrant three we have the following:

\[\cos(240^{\circ}) = -\cos(60^{\circ}) = -\frac{1}{2}\]

Example 4

\[\cos(120^{\circ})+2\sin^2(60^{\circ})-\tan^2(30^{\circ})\]

Solution:

\[\begin{align*} \cos(120^{\circ}) & = - \cos(60^{\circ}) = -\frac{1}{2}\\ \sin(60^{\circ}) & = \frac{\sqrt3}{2}\\ \tan(30^{\circ}) & = \frac{\sin(30^{\circ})}{\cos(30^{\circ})}\\ & = \frac{\frac12}{\frac{\sqrt3}{2}}\\ & = \frac{1}{\sqrt3} \end{align*}\]

\[\begin{align*} \cos(120^{\circ})+2\sin^2(60^{\circ})-\tan^2(30^{\circ}) & = \left(-\frac{1}{2}\right)+2\left(\frac{\sqrt3}{2}\right)^2-\left(\frac{1}{\sqrt3}\right)^2\\ & = -\frac{1}{2}+\frac{3}{2}-\frac{1}{3}\\ & = 1-\frac13\\ & = \frac{2}{3} \end{align*}\]

Example 5

\[\sin^2(45^{\circ})+3\cos^2(135^{\circ})-2\tan(225^{\circ})\]

Solution:

\[\begin{align*} \sin(45^{\circ}) & = \frac{\sqrt2}{2}\\ \cos(135^{\circ}) & = -\cos(45^{\circ}) = -\frac{\sqrt{2}}{2}\\ \tan(225^{\circ}) & = \frac{\sin(225)}{\cos(225)}=\frac{\sin(45)}{\cos(45)}\\ & = \frac{\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=1 \end{align*}\]

\[\begin{align*} \sin^2(45^{\circ})+3\cos^2(135^{\circ})-2\tan(225^{\circ}) & = \left(\frac{\sqrt2}{2}\right)^2+3\left(-\frac{\sqrt2}{2}\right)^2-2(1)\\ & = \frac{1}{2}+3\left(\frac{1}{2}\right)-2\\ & = 2-2 =0 \end{align*}\]

Example 6

If \(\cos(\theta) = \frac{\sqrt3}{2}\) and \(270^{\circ}<\theta<360^{\circ}\), find \(\theta\).

Solution:

We know that \(\cos(30^{\circ})=\frac{\sqrt3}{2}\). However, \(\theta\) is in fourth quadrant. Therefore,

\[\theta = 360^{\circ} - 30^{\circ} =330^{\circ}\]

Example 7

If \(\sin(\theta) = \frac{\sqrt2}{2}\) and \(90^{\circ}<\theta<180^{\circ}\), find \(\theta\).

Solution:

We know that \(\sin(45^{\circ})=\frac{\sqrt2}{2}\) and \(\theta\) is in the second quadrant. Therefore,

\[\theta = 180^{\circ} - 45^{\circ} = 135^{\circ}\]

Example 8

If \(\tan(\theta) = \frac{\sqrt3}{3}\) and \(180^{\circ}<\theta<270^{\circ}\), find \(\theta\).

Solution:

First, it may help to rationalize the numerator

\[ \dfrac{\sqrt{3}}{3}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{3}{3\sqrt{3}}=\dfrac{1}{\sqrt{3}}. \]

Next, divide the top and bottom by \(2\)

\[ \dfrac{1}{\sqrt{3}}\cdot\dfrac{\frac{1}{2}}{\frac{1}{2}}=\dfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}. \]

Since, \(\tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta)}\) we want to find a \(\theta\) such that \(\cos(\theta)=\frac{\sqrt{3}}{2}\) and \(\sin(\theta)=\dfrac{1}{2}\). Recall, \(\cos(30)=\frac{\sqrt{3}}{2}\) and \(\sin(30)=\frac{1}{2}\). Since \(\theta\) is in quadrant III we know

\[ \theta=180+30=210^{\circ} \]