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Section 1.3 Unit Circle

In this section, we will introduce the trigonometric functions using the Unit Circle.

Subsection 1.3.1 Unit Circle

Definition 1.3.1. Unit Circle.

The unit circle is a circle whose radius is 1 and whose center is at the origin of a rectangular plane (or \(xy\)-plane). The equation for the unit circle is
\begin{equation*} x^2+y^2=1 \end{equation*}
Let \(t\) be a real number. Recall from Definition 1.2.18 that a radian measure of a central angle, \(t\text{,}\) is defined as the ratio of the arc length \(s\) to the radius \(r\text{.}\) In other words, \(t=\frac{s}{r}\text{.}\) In the unit circle, the radius is \(r=1\text{,}\) and the angle in radians is equal to the arc length, \(t=s\text{.}\) We will let \(t\) be in radians. The circumference of the unit circle is \(2\pi r=2\pi \cdot 1=2\pi\text{.}\)
If \(t\geq0\text{,}\) we can imagine wrapping a line around the unit circle, marking off a distance of \(t\) in a counterclockwise direction, and labeling that point \(P(x,y)\text{,}\) whic becomes the terminal point. If \(t\lt 0\) then we would wrap in a clockwise direction.
If \(t\gt2\pi\) or \(t\lt -2\pi\text{,}\) then the length is longer than the circumference of the unit circle and you will need to travel around the unit circle more than once before arrive at the point \(P(x,y)\text{.}\) Therefore, we can conclude that regardless of the value of \(t\text{,}\) we have a unique point \(P(x,y)\) that lies on the unit circle. We call \(P(x,y)\) the point on the unit circle that corresponds to \(t\).

Subsection 1.3.2 Trigonometric Functions

The \(x\)- and \(y\)-coordinates for \(P(x,y)\) can then be used to define the six trigonometric functions of a real number \(t\text{:}\)
sine cosine tangent cosecant secant cotangent
which are abbreviated as sin, cos, tan, csc, sec, and cot, respectively.

Definition 1.3.2. Definition of Trigonometric Functions.

Let \(t\) be any real number and let \(P(x,y)\) be the terminal point on the unit circle associated with \(t\text{.}\) Then
\begin{align*} \sin t\amp =y \amp \cos t\amp =x \amp \tan t\amp =\frac{y}{x},~(x\neq0)\\ \csc t\amp =\frac{1}{y},~(y\neq0) \amp \sec t\amp =\frac{1}{x},~(x\neq0) \amp \cot t\amp =\frac{x}{y},~(y\neq0) \end{align*}
Notice that \(\tan t\) and \(\sec t\) re undefined when \(x=0\) and \(\csc t\) and \(\cot t\) are undefined when \(y=0\text{.}\)

Example 1.3.3.

Let \(t\) be the angle that corresponds to the point \(P(\frac{\sqrt{3}}{2},-\frac{1}{2})\text{.}\) Find the exact values of the six trigonometric functions corresponding to \(t\text{:}\) \(\sin t\text{,}\) \(\cos t\text{,}\) \(\tan t\text{,}\) \(\csc t\text{,}\) \(\sec t\text{,}\) \(\cot t\text{.}\)

Solution.

The point \(P(\frac{\sqrt{3}}{2},-\frac{1}{2})\) gives us \(x=\frac{\sqrt{3}}{2}\) and \(y=-\frac{1}{2}\text{.}\) Then we have
\begin{align*} \sin \theta\amp =y=-\frac{1}{2},\amp \csc \theta\amp =\frac{1}{y}=\frac{1}{-\frac{1}{2}}=-2,\\ \cos \theta\amp =x=\frac{\sqrt{3}}{2}, \amp \sec \theta\amp =\frac{1}{x}=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3},\\ \tan \theta\amp =\frac{y}{x}=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3},\amp \cot \theta\amp =\frac{x}{y}=\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=-\sqrt{3}. \end{align*}

Subsection 1.3.3 Trigonometric Functions of an Angle

Definition 1.3.4. Trigonometric Functions of an Angle.

If \(\theta\) is an angle with radian measure \(t\text{,}\) then the six trigonometric functions become
\begin{align*} \sin \theta\amp =y \amp \cos \theta\amp =x \amp \tan \theta\amp =\frac{y}{x},~(x\neq0)\\ \csc \theta\amp =\frac{1}{y},~(y\neq0) \amp \sec \theta\amp =\frac{1}{x},~(x\neq0) \amp \cot \theta\amp =\frac{x}{y},~(y\neq0) \end{align*}

Example 1.3.5.

Find the exact values of the six trigonometric functions for
(a)
\(\theta=0\)
Solution.
When \(\theta=0\) radians (\(0^{\circ}\)), the point on the circle is \(P(1,0)\text{.}\)
Then \(x=1\) and \(y=0\) gives us
\begin{align*} \sin 0\amp =\sin 0^{\circ}=0, \amp \csc 0\amp =\csc 0^{\circ}=\mbox{undefined},\\ \cos 0\amp =\cos 0^{\circ}=1, \amp \sec 0\amp =\sec 0^{\circ}=1,\\ \tan 0\amp =\tan 0^{\circ}=0, \amp \cot 0 \amp =\cot 0^{\circ}=\mbox{undefined}. \end{align*}
(b)
\(\theta=\frac{3\pi}{2}\)
Solution.
When \(\theta=\frac{3\pi}{2}\) radians (\(270^{\circ}\)), the point on the circle is \(P(0,-1)\text{.}\)
Then \(x=0\) and \(y=-1\) gives us
\begin{align*} \sin \frac{3\pi}{2}\amp =\sin 270^{\circ}=-1, \amp \csc \frac{3\pi}{2}\amp =\csc 270^{\circ}=-1,\\ \cos \frac{3\pi}{2}\amp =\cos 270^{\circ}=0, \amp \sec \frac{3\pi}{2}\amp =\sec 270^{\circ}=\mbox{undefined},\\ \tan \frac{3\pi}{2}\amp =\tan 270^{\circ}=\mbox{undefined}, \amp \cot \frac{3\pi}{2}\amp =\cot 270^{\circ}=0 \end{align*}
(c)
\(\theta=5\pi\)
Solution.
Since \(\theta=5\pi>2\pi\text{,}\) our angle is greater than one full rotation of a circle. We first subtract \(\theta\) by one rotation, \(2\pi\text{,}\) to get
\begin{equation*} 5\pi-2\pi-=3\pi \end{equation*}
Once again, since we have completed more than one full rotation, we can repeat the previous step:
\begin{equation*} 3\pi-2\pi=\pi \end{equation*}
The values of the six trigonometric functions when \(\theta=5\pi\) are equal to those when \(\theta=\pi\text{.}\) Notice that \(5\pi\) and \(\pi\) are coterminal angles, both ending at the point\(P(-1,0)\text{.}\)
Since \(x=-1\) and \(y=0\) we have
\begin{align*} \sin 5\pi\amp =0, \amp \cos 5\pi\amp =-1, \amp \tan 5\pi\amp =0,\\ \csc 5\pi\amp =\mbox{undefined}, \amp \sec 5\pi\amp =-1, \amp \cot 5\pi\amp =\mbox{undefined}. \end{align*}

Example 1.3.6. Finding the Exact Values of the Trigonometric Functions for \(\theta=45^{\circ}\).

Find the exact values of the six trigonometric functions for \(\theta=45^{\circ}\text{.}\)

Solution.

We begin by drawing a right triangle with a base angle of \(45^{\circ}\) in the unit circle.
Since the first quadrant has \(90^{\circ}\text{,}\) at \(\theta=45^{\circ}\text{,}\) the point \(P\) lies on the line that bisects the first quadrant. This means the point \(P\) is on the line \(y=x\text{.}\) Since \(P(x,y)\) also lies on the unit circle, whose equation is \(x^2+y^2=1\text{,}\) we get
\begin{align*} x^2+y^2\amp = 1\\ x^2+x^2\amp = 1 \amp(\mbox{since } y=x)\\ 2x^2\amp = 1\\ x^2\amp = \frac{1}{2}\\ x\amp = \frac{1}{\sqrt{2}}\\ y\amp = \frac{1}{\sqrt{2}} \amp(\mbox{since } y=x) \end{align*}
Then
\begin{align*} \sin 45^{\circ} \amp=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \amp \csc 45^{\circ}=\frac{1}{\frac{\sqrt{2}}{2}}\amp=\sqrt{2}\\ \cos 45^{\circ} \amp=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \amp \sec 45^{\circ}=\frac{1}{\frac{\sqrt{2}}{2}}\amp=\sqrt{2}\\ \tan 45^{\circ} \amp=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=1 \amp \cot 45^{\circ}=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\amp=1 \end{align*}

Example 1.3.7. Finding the Exact Values of the Trigonometric Functions for \(\theta=30^{\circ}\).

Find the exact values of the six trigonometric functions for \(\theta=30^{\circ}\text{.}\)

Solution.

First, we will draw a triangle in a circle with an angle of \(30^{\circ}\) and a second triangle with an angle of \(-30^{\circ}\) .
This gives us two 30-60-90 triangles. Notice this now gives us one larger triangle whose angles are all \(60^{\circ}\text{.}\) Thus we have an equilateral triangle, with each side of length 1.
We see that \(1=2y\) so \(y=\frac{1}{2}\text{.}\) Then by the Pythagorean Theorem,
\begin{align*} x^2+y^2\amp = 1^2\\ x^2+\left(\frac{1}{2}\right)^2\amp = 1\\ x^2+\frac{1}{4}\amp = 1\\ x^2\amp = \frac{3}{4}\\ x\amp = \frac{\sqrt{3}}{2} \end{align*}
Giving us the following triangle
Then
\begin{align*} \sin 30^{\circ}\amp =\frac{1}{2}, \amp \csc 30^{\circ}\amp =\frac{1}{\frac{1}{2}}=2,\\ \cos 30^{\circ}\amp =\frac{\sqrt{3}}{2}, \amp \sec 30^{\circ}\amp =\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3},\\ \tan 30^{\circ}\amp =\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}, \amp \cot 30^{\circ}\amp =\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}. \end{align*}

Remark 1.3.8. Finding the Exact Values of the Trigonometric Functions for \(\theta=90^{\circ}\).

Similarly, we can get the following for \(\theta=60^{\circ}\text{.}\)
We now summarize what we know about the six trigonometric functions for special angles. Note the trigonometric functions for \(\theta=\frac{\pi}{2}\) and \(\theta=\frac{\pi}{3}\) are left as exercises.
Table 1.3.9. Trigonometric functions for special angles
\(\theta\) (deg) \(\theta\) (rad) \(\sin\theta\) \(\cos\theta\) \(\tan\theta\) \(\csc\theta\) \(\sec\theta\) \(\cot\theta\)
\(0^{\circ}\) 0 0 1 0 undef 1 undef
\(30^{\circ}\) \(\dfrac{\pi}{6}\) \(\dfrac{1}{2}\) \(\dfrac{\sqrt{3}}{2}\) \(\dfrac{\sqrt{3}}{3}\) 2 \(\dfrac{2\sqrt{3}}{3}\) \(\sqrt{3}\)
\(45^{\circ}\) \(\dfrac{\pi}{4}\) \(\dfrac{\sqrt{2}}{2}\) \(\dfrac{\sqrt{2}}{2}\) 1 \(\sqrt{2}\) \(\sqrt{2}\) 1
\(60^{\circ}\) \(\dfrac{\pi}{3}\) \(\dfrac{\sqrt{3}}{2}\) \(\dfrac{1}{2}\) \(\sqrt{3}\) \(\dfrac{2\sqrt{3}}{3}\) 2 \(\dfrac{\sqrt{3}}{3}\)
\(90^{\circ}\) \(\dfrac{\pi}{2}\) 1 0 undef 1 undef 0

Subsection 1.3.4 Symmetry on the Unit Circle

If the point \(P(x,y)\) lies on the unit circle, the following symmetric points also lie on the unit circle:
  1. \(Q(-x,y)\text{:}\) Symmetry about the \(y\)-axis.
  2. \(R(-x,-y)\text{:}\) Symmetry about the origin.
  3. \(S(x,-y)\text{:}\) Symmetry about the \(x\)-axis.
This symmetry within the unit circle resembles the pattern observed in the Star Compass. When a star emerges in the eastern sky, it will eventually descend and set in the corresponding house of the western sky. For instance, if a star rises above the horizon in the Nālani house of the Ko‘olau quadrant (northeast), it will journey across the sky and set in the equivalent house within the Ho‘olua quadrant (northwest). This similarity aligns with the symmetry between points \(P(x,y)\) and \(Q(-x,y)\text{.}\) Additionally, if an ocean swell or wind originates from the Nālani house in the Malanai quadrant (southeast), it will pass the wa‘a and exit in the opposite direction toward the Ho‘olua quadrant (northwest), still within the Nālani house. This mirrors the symmetry between points \(S(x,-y)\) and \(Q(-x,y)\text{.}\)
A fourth form of symmetry involves reflecting points across the diagonal line \(y=x\text{,}\) where the \(x\)- and \(y\)-values are equal.
  1. \(T(y,x)\text{:}\) Symmetry about the line \(y=x\text{.}\) This is accomplished by interchanging the \(x\)- and \(y\)-values.
Notice on the Unit Circle that the radius extending from the center at an angle of \(30^{\circ}\) to the point \(T(x,y)=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\) is symmetric about the line \(y=x\text{,}\) in relation to the radius extending from the center at an angle of \(60^{\circ}\) to the point \(P(x,y)\text{.}\)
Using symmetry about the \(x\)-axis, symmetry about the \(y\)-axis, and symmetry about the origin, we can complete the unit circle, as long as we remember that the \(x\)-values in Quadrants II and III are negative while the \(y\)-values in Quadrants III and IV are negative.
Finally, we tie everything together and look at the entire Unit Circle. At first glance it may seem intimidating, however, similar to the Star Compass, there is a lot of symmetry (\(x\)-axis, \(y\)-axis, origin, about the line \(y=x\)) and it can help by focusing on one quadrant, and use symmetry to fill out the rest of the circle.
Figure 1.3.10. The Unit Circle for common angles in radians and degrees.

Subsection 1.3.5 Trigonometric Functions on a Circle with Radius \(r\)

Until now, computing the exact values of trigonometric functions of an angle \(\theta\) required us to locate the corresponding point \(P(x,y)\) on the unit circle. However, we can use any circle with center at the origin, that is, any circle of the form \(x^2+y^2=r^2\text{,}\) where \(r>0\) is the radius. Note that if \(r=1\text{,}\) then it is the unit circle.

Exercises 1.3.6 Exercises

Exercise Group.

Verify algebraically that the point \(P\) is on the unit circle (\(x^2+y^2=1\))
1.
\(P\left(\frac{3}{5},-\frac{4}{5}\right)\)
Answer.
\(\left(\frac{3}{5}\right)^2+\left(-\frac{4}{5}\right)^2=1\)
2.
\(P\left(-\frac{\sqrt{39}}{8},-\frac{5}{8}\right)\)
Answer.
\(\left(-\frac{\sqrt{39}}{8}\right)^2+\left(-\frac{5}{8}\right)^2=1\)
3.
\(P\left(-\frac{\sqrt{55}}{8},\frac{3}{8}\right)\)
Answer.
\(\left(-\frac{\sqrt{55}}{8}\right)^2+\left(\frac{3}{8}\right)^2=1\)
4.
\(P\left(-\frac{2}{3},\frac{\sqrt{5}}{3}\right)\)
Answer.
\(\left(-\frac{2}{3}\right)^2+\left(\frac{\sqrt{5}}{3}\right)^2=1\)
5.
\(P\left(\frac{3}{4},\frac{\sqrt{7}}{4}\right)\)
Answer.
\(\left(\frac{3}{4}\right)^2+\left(\frac{\sqrt{7}}{4}\right)^2=1\)
6.
\(P\left(\frac{\sqrt{21}}{5},-\frac{2}{5}\right)\)
Answer.
\(\left(\frac{\sqrt{21}}{5}\right)^2+\left(-\frac{2}{5}\right)^2=1\)

Exercise Group.

Let the point \(P\) be on the unit circle. Given the quadrant that \(P\) lies in, determine the missing coordinate, \(a\)
7.
III; \(P\left(-\frac{2}{3},a\right)\)
Answer.
\(-\frac{\sqrt{5}}{3}\)
8.
IV; \(P\left(\frac{5}{8},a\right)\)
Answer.
\(-\frac{\sqrt{39}}{8}\)
9.
III; \(P\left(a,-\frac{2}{5}\right)\)
Answer.
\(-\frac{\sqrt{21}}{5}\)
10.
II; \(P\left(a,\frac{4}{9}\right)\)
Answer.
\(-\frac{\sqrt{65}}{9}\)

Exercise Group.

Given an angle \(\theta\) that corresponds to the point \(P\) on the unit circle, determine the coordinates of the point \(P(x,y)\text{.}\)
11.
\(\theta=\frac{\pi}{2}\)
Answer.
\((0,1)\)
12.
\(\theta=\pi\)
Answer.
\((-1,0)\)
13.
\(\theta=\frac{5\pi}{3}\)
Answer.
\(\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)\)
14.
\(\theta=\frac{4\pi}{3}\)
Answer.
\(\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)\)
15.
\(\theta=-\frac{\pi}{4}\)
Answer.
\(\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)\)
16.
\(\theta=\frac{5\pi}{6}\)
Answer.
\(\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\)
17.
\(\theta=315^{\circ}\)
Answer.
\(\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)\)
18.
\(\theta=720^{\circ}\)
Answer.
\(\left(1,0\right)\)
19.
\(\theta=60^{\circ}\)
Answer.
\(\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\)
20.
\(\theta=-180^{\circ}\)
Answer.
\(\left(-1,0\right)\)
21.
\(\theta=210^{\circ}\)
Answer.
\(\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\)
22.
\(\theta=120^{\circ}\)
Answer.
\(\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\)

Exercise Group.

For each angle \(\theta\) in Exercises 1.3.6.11–22, find the exact values of the six trigonometric functions. If any are not defined, say “undefined.”
23.
\(\theta=\frac{\pi}{2}\)
Answer.
\(\sin\frac{\pi}{2}=1\text{;}\) \(\cos\frac{\pi}{2}=0\text{;}\) \(\tan\frac{\pi}{2}\) is undefined; \(\csc\frac{\pi}{2}=1\text{;}\) \(\sec\frac{\pi}{2}\) is undefined; \(\cot\frac{\pi}{2}=0\)
24.
\(\theta=\pi\)
Answer.
\(\sin\pi=0\text{;}\) \(\cos\pi=-1\text{;}\) \(\tan\pi=0\text{;}\) \(\csc\pi\) is undefined; \(\sec\pi=-1\text{;}\) \(\cot\pi\) is undefined
25.
\(\theta=\frac{5\pi}{3}\)
Answer.
\(\sin\frac{5\pi}{3}=-\frac{\sqrt{3}}{2}\text{;}\) \(\cos\frac{5\pi}{3}=\frac{1}{2}\text{;}\) \(\tan\frac{5\pi}{3}=-\sqrt{3}\text{;}\) \(\csc\frac{5\pi}{3}=-\frac{2\sqrt{3}}{3}\text{;}\) \(\sec\frac{5\pi}{3}=2\text{;}\) \(\cot\frac{5\pi}{3}=-\frac{\sqrt{3}}{3}\)
26.
\(\theta=\frac{4\pi}{3}\)
Answer.
\(\sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2}\text{;}\) \(\cos\frac{4\pi}{3}=-\frac{1}{2}\text{;}\) \(\tan\frac{4\pi}{3}=\sqrt{3}\text{;}\) \(\csc\frac{4\pi}{3}=-\frac{2\sqrt{3}}{3}\text{;}\) \(\sec\frac{4\pi}{3}=-2\text{;}\) \(\cot\frac{4\pi}{3}=\frac{\sqrt{3}}{3}\)
27.
\(\theta=-\frac{\pi}{4}\)
Answer.
\(\sin\left(-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\text{;}\) \(\cos\left(-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\text{;}\) \(\tan\left(-\frac{\pi}{4}\right)=-1\text{;}\) \(\csc\left(-\frac{\pi}{4}\right)=-\sqrt{2}\text{;}\) \(\sec\left(-\frac{\pi}{4}\right)=\sqrt{2}\text{;}\) \(\cot\left(-\frac{\pi}{4}\right)=-1\)
28.
\(\theta=\frac{5\pi}{6}\)
Answer.
\(\sin\frac{5\pi}{6}=\frac{1}{2}\text{;}\) \(\cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}\text{;}\) \(\tan\frac{5\pi}{6}=-\frac{\sqrt{3}}{3}\text{;}\) \(\csc\frac{5\pi}{6}=2\text{;}\) \(\sec\frac{5\pi}{6}=-\frac{2\sqrt{3}}{3}\text{;}\) \(\cot\frac{5\pi}{6}=-\sqrt{3}\)
29.
\(\theta=315^{\circ}\)
Answer.
\(\sin315^{\circ}=-\frac{\sqrt{2}}{2}\text{;}\) \(\cos315^{\circ}=\frac{\sqrt{2}}{2}\text{;}\) \(\tan315^{\circ}=-1\text{;}\) \(\csc315^{\circ}=-\sqrt{2}\text{;}\) \(\sec315^{\circ}=\sqrt{2}\text{;}\) \(\cot315^{\circ}=-1\)
30.
\(\theta=720^{\circ}\)
Answer.
\(\sin720^{\circ}=0\text{;}\) \(\cos720^{\circ}=1\text{;}\) \(\tan720^{\circ}=0\text{;}\) \(\csc720^{\circ}\) is undefined; \(\sec720^{\circ}=1\text{;}\) \(\cot720^{\circ}\) is undefined
31.
\(\theta=60^{\circ}\)
Answer.
\(\sin60^{\circ}=\frac{\sqrt{3}}{2}\text{;}\) \(\cos60^{\circ}=\frac{1}{2}\text{;}\) \(\tan60^{\circ}=\sqrt{3}\text{;}\) \(\csc60^{\circ}=\frac{2\sqrt{3}}{3}\text{;}\) \(\sec60^{\circ}=2\text{;}\) \(\cot60^{\circ}=\frac{\sqrt{3}}{3}\)
32.
\(\theta=-180^{\circ}\)
Answer.
\(\sin(-180^{\circ})=0\text{;}\) \(\cos(-180^{\circ})=-1\text{;}\) \(\tan(-180^{\circ})=0\text{;}\) \(\csc(-180^{\circ})\) is undefined; \(\sec(-180^{\circ})=-1\text{;}\) \(\cot(-180^{\circ})\) is undefined
33.
\(\theta=210^{\circ}\)
Answer.
\(\sin210^{\circ}=-\frac{1}{2}\text{;}\) \(\cos210^{\circ}=-\frac{\sqrt{3}}{2}\text{;}\) \(\tan210^{\circ}=\frac{\sqrt{3}}{3}\text{;}\) \(\csc210^{\circ}=-2\text{;}\) \(\sec210^{\circ}=-\frac{2\sqrt{3}}{3}\text{;}\) \(\cot210^{\circ}=\sqrt{3}\)
34.
\(\theta=120^{\circ}\)
Answer.
\(\sin120^{\circ}=\frac{\sqrt{3}}{2}\text{;}\) \(\cos120^{\circ}=-\frac{1}{2}\text{;}\) \(\tan120^{\circ}=-\sqrt{3}\text{;}\) \(\csc120^{\circ}=\frac{2\sqrt{3}}{3}\text{;}\) \(\sec120^{\circ}=-2\text{;}\) \(\cot120^{\circ}=-\frac{\sqrt{3}}{3}\)

Exercise Group.

Let \(\theta\) be the angle that corresponds to the point \(P\text{.}\) Exercises 1.3.6.1–6 verified \(P\) is on the unit circle. Find the exact values of the six trigonometric functions of \(\theta\text{.}\)
35.
\(P\left(\frac{3}{5},-\frac{4}{5}\right)\)
Answer.
\(\sin\theta=-\frac{4}{5}\text{;}\) \(\cos\theta=\frac{3}{5}\text{;}\) \(\tan\theta=-\frac{4}{3}\text{;}\) \(\csc\theta=-\frac{5}{4}\text{;}\) \(\sec\theta=\frac{5}{3}\text{;}\) \(\cot\theta=-\frac{3}{4}\)
36.
\(P\left(-\frac{\sqrt{39}}{8},-\frac{5}{8}\right)\)
Answer.
\(\sin\theta=-\frac{5}{8}\text{;}\) \(\cos\theta=-\frac{\sqrt{39}}{8}\text{;}\) \(\tan\theta=\frac{5\sqrt{39}}{39}\text{;}\) \(\csc\theta=-\frac{8}{5}\text{;}\) \(\sec\theta=-\frac{8\sqrt{39}}{39}\text{;}\) \(\cot\theta=\frac{\sqrt{39}}{5}\)
37.
\(P\left(-\frac{\sqrt{55}}{8},\frac{3}{8}\right)\)
Answer.
\(\sin\theta=\frac{3}{8}\text{;}\) \(\cos\theta=-\frac{\sqrt{55}}{8}\text{;}\) \(\tan\theta=-\frac{3\sqrt{55}}{55}\text{;}\) \(\csc\theta=\frac{8}{3}\text{;}\) \(\sec\theta=-\frac{8\sqrt{55}}{55}\text{;}\) \(\cot\theta=-\frac{\sqrt{55}}{3}\)
38.
\(P\left(-\frac{2}{3},\frac{\sqrt{5}}{3}\right)\)
Answer.
\(\sin\theta=\frac{\sqrt{5}}{3}\text{;}\) \(\cos\theta=-\frac{2}{3}\text{;}\) \(\tan\theta=-\frac{\sqrt{5}}{2}\text{;}\) \(\csc\theta=\frac{3\sqrt{5}}{5}\text{;}\) \(\sec\theta=-\frac{3}{2}\text{;}\) \(\cot\theta=-\frac{2\sqrt{5}}{5}\)
39.
\(P\left(\frac{3}{4},\frac{\sqrt{7}}{4}\right)\)
Answer.
\(\sin\theta=\frac{\sqrt{7}}{4}\text{;}\) \(\cos\theta=\frac{3}{4}\text{;}\) \(\tan\theta=\frac{\sqrt{7}}{3}\text{;}\) \(\csc\theta=\frac{4\sqrt{7}}{7}\text{;}\) \(\sec\theta=\frac{4}{3}\text{;}\) \(\cot\theta=\frac{3\sqrt{7}}{7}\)
40.
\(P\left(\frac{\sqrt{21}}{5},-\frac{2}{5}\right)\)
Answer.
\(\sin\theta=-\frac{2}{5}\text{;}\) \(\cos\theta=\frac{\sqrt{21}}{5}\text{;}\) \(\tan\theta=-\frac{2\sqrt{21}}{21}\text{;}\) \(\csc\theta=-\frac{5}{2}\text{;}\) \(\sec\theta=\frac{5\sqrt{21}}{21}\text{;}\) \(\cot\theta=-\frac{\sqrt{21}}{2}\)

Exercise Group.

Find the exact value of each expression.
41.
\(\sin30^{\circ}+\sin150^{\circ}\)
Answer.
\(1\)
42.
\(\cos30^{\circ}+\cos150^{\circ}\)
Answer.
\(0\)
43.
\(\sin60^{\circ}+\sin120^{\circ}+\sin240^{\circ}+\sin300^{\circ}\)
Answer.
\(0\)
44.
\(\cos60^{\circ}+\cos120^{\circ}+\cos240^{\circ}+\cos300^{\circ}\)
Answer.
\(0\)
45.
\(\tan45^{\circ}+\tan135^{\circ}\)
Answer.
\(0\)
46.
\(\tan135^{\circ}+\tan225^{\circ}\)
Answer.
\(0\)
47.
\(\tan225^{\circ}+\tan315^{\circ}\)
Answer.
\(0\)
48.
\(\tan45^{\circ}+\tan225^{\circ}\)
Answer.
\(2\)