Supplemental Videos.
The main topics of this section are also presented in the following videos:
Section 1.3 The Sine and Cosine Functions
In the previous section, we found Cartesian coordinates of points on the unit circle corresponding to common angles. Next, we explore how these \((x,y)\) coordinates are related to periodic functions.
Subsection 1.3.1 The Definition of Sine and Cosine
The Sine and Cosine Functions.
Given an angle \(\theta \) (in either degrees or radians) and the \((x,y)\) coordinates of the corresponding point on the unit circle, we define cosine and sine as
\begin{equation*}
\cos(\theta)=x \hspace{.25in} \text{and} \hspace{.25in} \sin(\theta)=y
\end{equation*}
Note that sine and cosine are functions that take angles as inputs.
unit circle with (x,y) and theta
Using the \((x,y)\) coordinates that we found in the last section, we can find the cosine and sine values of common angles on the unit circle.
Example 1.24.
Find \(\cos(30^\circ)\) and \(\sin(30^\circ)\text{.}\)
Solution.
Since \(30^\circ\) is a common angle on the unit circle, we have that
\begin{equation*}
\cos(30^\circ)=\frac{\sqrt{3}}{2} \hspace{.25in} \text{and} \hspace{.25in} \sin(30^\circ)=\frac{1}{2}
\end{equation*}
Example 1.25.
Find \(\displaystyle \cos\left(\frac{5\pi}{4}\right)\) and \(\displaystyle \sin\left(\frac{5\pi}{4}\right)\text{.}\)
Solution.
Since \(\displaystyle \frac{5\pi}{4}\) is a common angle on the unit circle, we have that
\begin{equation*}
\cos\left(\frac{5\pi}{4}\right)=-\frac{\sqrt{2}}{2} \hspace{.25in} \text{and} \hspace{.25in} \sin\left(\frac{5\pi}{4}\right)=-\frac{\sqrt{2}}{2}
\end{equation*}
Example 1.26.
Find \(\cos(\theta)\) and \(\sin(\theta)\) if \(\theta = \pi\text{.}\)
Solution.
\begin{equation*}
\cos(\pi)=-1 \hspace{.25in} \text{and} \hspace{.25in} \sin(\pi)=0
\end{equation*}
The special values of sine and cosine in the first quadrant are very useful to know, since knowing them allows you to quickly evaluate the sine and cosine of common angles without needing to look at a reference sheet or use your calculator. However, scenarios do come up where we need to know the sine and cosine of other angles.
To find the cosine and sine of angles that are not common angles on the unit circle, we can use a calculator or a computer.
Example 1.27.
Find \(\cos(20^\circ)\) and \(\sin(20^\circ)\text{.}\)
Solution.
Since \(20^\circ\) is not a common angle on the unit circle, we need to use a calculator to find approximate values for \(\cos(20^\circ)\) and \(\sin(20^\circ)\text{.}\) Checking that our calculator is in degree mode, we get that
\begin{equation*}
\cos(20^\circ)\approx 0.9397 \hspace{.25in} \text{and} \hspace{.25in} \sin(20^\circ)\approx 0.3420
\end{equation*}
Remember that these values correspond to the \(x\) and \(y\) coordinates of the point on the unit circle corresponding to an angle of 20 degrees.
unit circle with (x,y) and theta at 20 degrees
Caution 1.28.
Most calculators can be set into “degree” or “radian” mode, which tells the calculator the units for the input value. When you evaluate "\(\cos(20)\)" on your calculator, it will evaluate it as the cosine of 20 degrees if the calculator is in degree mode or the cosine of 20 radians if the calculator is in radian mode. Always make sure to check what mode your calculator is in when working with trigonometric functions!
Example 1.29.
Find \(\cos(2.5)\) and \(\sin(2.5)\text{.}\)
Solution.
Since there is no degree symbol after the number 2.5, we assume that 2.5 is in radians. Since 2.5 radians is not a common angle on the unit circle, we need to use a calculator to find approximate values for \(\cos(2.5)\) and \(\sin(2.5)\text{.}\) Checking that our calculator is in radian mode, we get that
\begin{equation*}
\cos(2.5)\approx -0.8011 \hspace{.25in} \text{and} \hspace{.25in} \sin(2.5)\approx 0.5985
\end{equation*}
Remember that these values correspond to the \(x\) and \(y\) coordinates of the point on the unit circle corresponding to an angle of 2.5 radians. To determine approximately where this is on the unit circle, we can find decimal approximations of a few common angles on the unit circle. Since \(\pi/2\) is approximately 1.57 and \(\pi\) is approximately 3.14, 2.5 radians is in between \(\pi/2\) radians and \(\pi\) radians. Therefore, 2.5 radians corresponds to a point on the unit circle in the second quadrant, and it makes sense that the cosine value or \(x\) coordinate is negative and the sine value or \(y\) coordinate is positive.
unit circle with (x,y) and theta at 2.5 radians
Subsection 1.3.2 Right Triangle Relationships
So far we have defined cosine and sine as the \(x\) and \(y\) coordinates of points on the unit circle. We can also state equivalent, but more general definitions of sine and cosine using a right triangle. On the right triangle, we label the hypotenuse as well as the side opposite the angle and the side adjacent (next to) the angle.
Given a right triangle with an angle of \(\theta\text{,}\) we define sine and cosine as
\begin{align*}
\sin(\theta) \amp =\frac{\text{opposite}}{\text{hypotenuse}} \\
\\
\cos(\theta) \amp =\frac{\text{adjacent}}{\text{hypotenuse}}
\end{align*}
right triangle with labels
A common mnemonic for remembering these relationships is Soh-Cah-Toa formed by the first letters of "Sine is opposite over hypotenuse" and "Cosine is adjacent over hypotenuse." We will return to "Toa" in the next section.
Example 1.30.
Given the triangle shown below, find the value for \(\cos(\theta)\text{.}\)
right triangle example
Solution.
The side adjacent to angle \(\theta\) is 15, and the hypotenuse of the triangle is 17. Using the definition of cosine given above, we get that
\begin{equation*}
\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{15}{17}
\end{equation*}
When working with general right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the sine and cosine of either of the two acute angles in any right triangle.
rotated right triangle
Example 1.31.
Using the triangle shown below, find \(\cos(\alpha)\text{,}\) \(\sin(\alpha)\text{,}\) \(\cos(\beta)\text{,}\) and \(\sin(\beta)\text{.}\)
rotated right triangle example
Solution.
Before solving for the cosine and sine values of \(\alpha\) and \(\beta\text{,}\) we must first find the length of the hypotenuse. Using the Pythagorean theorem, we get that
\begin{align*}
c^2 \amp= 3^2 + 4^2 \\
c^2 \amp= 9 + 16 \\
c^2 \amp= 25 \\
c \amp = 5
\end{align*}
rotated right triangle example with hypotenuse labeled
Now solving for the cosine and sine values of \(\alpha\) and \(\beta\text{,}\) we get that
\begin{equation*}
\cos(\alpha) = \frac{\text{adjacent to } \alpha}{\text{hypotenuse}}=\frac{3}{5} \hspace{.5in} \cos(\beta) = \frac{\text{adjacent to } \beta}{\text{hypotenuse}}=\frac{4}{5}
\end{equation*}
\begin{equation*}
\sin(\alpha) = \frac{\text{opposite to } \alpha}{\text{hypotenuse}}=\frac{4}{5} \hspace{.5in} \sin(\beta) = \frac{\text{opposite to } \beta}{\text{hypotenuse}}=\frac{3}{5}
\end{equation*}
Example 1.32.
A right triangle is drawn with angle \(\alpha\) opposite a side with length 33 and angle \(\beta\) opposite a side with length 56. Find the sine and cosine of \(\alpha\) and \(\beta\text{.}\)
Solution.
Before solving for the cosine and sine values of \(\alpha\) and \(\beta\text{,}\) we must first find the length of the hypotenuse. Using the Pythagorean theorem, we get that
\begin{align*}
c^2 \amp= 56^2 + 33^2 \\
c^2 \amp= 4225 \\
c \amp = 65
\end{align*}
rotated right triangle example with hypotenuse labeled
Now solving for the cosine and sine values of \(\alpha\) and \(\beta\text{,}\) we get that
\begin{equation*}
\cos(\alpha) = \frac{\text{adjacent to } \alpha}{\text{hypotenuse}}=\frac{56}{65} \hspace{.5in} \cos(\beta) = \frac{\text{adjacent to } \beta}{\text{hypotenuse}}=\frac{33}{65}
\end{equation*}
\begin{equation*}
\sin(\alpha) = \frac{\text{opposite to } \alpha}{\text{hypotenuse}}=\frac{33}{65} \hspace{.5in} \sin(\beta) = \frac{\text{opposite to } \beta}{\text{hypotenuse}}=\frac{56}{65}
\end{equation*}
We have now discussed two different ways to define the sine and cosine values of an angle. By revisiting our first definition, which uses the unit circle, and comparing this to the second definition, which uses right triangle relationships, we can prove that these two definitions are equivalent.
We first defined cosine and sine as the \(x\) and \(y\) coordinates of a point on the unit circle. If we use that point to make a right triangle, as shown below, the radius of the unit circle corresponds to the hypotenuse of the right triangle and the \(x\) and \(y\) coordinates correspond to the legs of the right triangle.
unit circle with right triangle
Using our right triangle relationships, we get that
\begin{equation*}
\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{r} \hspace{.3in} \text{and} \hspace{.3in} \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{r}
\end{equation*}
Since the radius of the unit circle is 1, these simplify to
\begin{equation*}
\cos(\theta) = \frac{x}{r} = \frac{x}{1} = x \hspace{.3in} \text{and} \hspace{.3in} \sin(\theta) = \frac{y}{r} = \frac{y}{1} = y
\end{equation*}
Subsection 1.3.3 The Graphs of Sine and Cosine
Since sine and cosine are functions, we can create tables of input and output values and use them to sketch the graphs of these functions.
Consider the function \(f(\theta)=\sin(\theta)\text{.}\) Below is a table showing input values of \(\theta\) and the corresponding output values of \(\sin(\theta)\text{.}\) Note that for each angle, we use the \(y\) value of the corresponding point on the unit circle to determine the output value of the sine function.
|
\begin{equation*}
\theta
\end{equation*}
|
\begin{equation*}
\ \ \ 0 \ \ \
\end{equation*}
|
\begin{equation*}
\ \ \frac{\pi}{6} \ \
\end{equation*}
|
\begin{equation*}
\ \ \frac{\pi}{4} \ \
\end{equation*}
|
\begin{equation*}
\ \ \frac{\pi}{3} \ \
\end{equation*}
|
\begin{equation*}
\ \ \frac{\pi}{2} \ \
\end{equation*}
|
\begin{equation*}
\ \ \frac{2\pi}{3} \ \
\end{equation*}
|
\begin{equation*}
\ \ \frac{3\pi}{4} \ \
\end{equation*}
|
\begin{equation*}
\ \frac{5\pi}{6} \
\end{equation*}
|
\begin{equation*}
\ \ \ \pi \ \ \
\end{equation*}
|
\begin{equation*}
\ \frac{7\pi}{6} \
\end{equation*}
|
\begin{equation*}
\frac{5\pi}{4}
\end{equation*}
|
\begin{equation*}
\frac{4\pi}{3}
\end{equation*}
|
\begin{equation*}
\ \frac{3\pi}{2} \
\end{equation*}
|
|
\begin{equation*}
\sin(\theta)
\end{equation*}
|
\begin{equation*}
\ 0 \
\end{equation*}
|
\begin{equation*}
\frac{1}{2}
\end{equation*}
|
\begin{equation*}
\frac{\sqrt{2}}{2}
\end{equation*}
|
\begin{equation*}
\frac{\sqrt{3}}{2}
\end{equation*}
|
\begin{equation*}
1
\end{equation*}
|
\begin{equation*}
\frac{\sqrt{3}}{2}
\end{equation*}
|
\begin{equation*}
\frac{\sqrt{2}}{2}
\end{equation*}
|
\begin{equation*}
\frac{1}{2}
\end{equation*}
|
\begin{equation*}
\ 0 \
\end{equation*}
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\begin{equation*}
-\frac{1}{2}
\end{equation*}
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\begin{equation*}
-\frac{\sqrt{2}}{2}
\end{equation*}
|
\begin{equation*}
-\frac{\sqrt{3}}{2}
\end{equation*}
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\begin{equation*}
-1
\end{equation*}
|
Using this table, we can sketch the graph of \(f(\theta)=\sin(\theta)\) as shown below. Recall that the horizontal axis of the graph corresponds to the input variable and the vertical axis corresponds to the output. Plotting more points gives the full shape of the sine function.
sine graph
Notice how the sine values are positive between 0 and \(\pi\text{,}\) which correspond to the values of sine in quadrants 1 and 2. Since the \(y\) values in quadrants 1 and 2 are positive, the sine values between 0 and \(\pi\) are positive. Also notice that the sine values are negative between \(\pi\) and \(2\pi\text{.}\) These values correspond to the \(y\) values in quadrants 3 and 4, which are negative.
Now consider the function \(g(\theta)=\cos(\theta)\text{.}\) Below is a table showing input values of \(\theta\) and output values of \(\cos(\theta)\text{.}\) Note that for each angle, we use the \(x\) value of the point on the unit circle to determine the output value of the cosine function.
|
\begin{equation*}
\theta
\end{equation*}
|
\begin{equation*}
\ \ \ 0 \ \ \
\end{equation*}
|
\begin{equation*}
\ \ \frac{\pi}{6} \ \
\end{equation*}
|
\begin{equation*}
\ \ \frac{\pi}{4} \ \
\end{equation*}
|
\begin{equation*}
\ \ \frac{\pi}{3} \ \
\end{equation*}
|
\begin{equation*}
\ \ \frac{\pi}{2} \ \
\end{equation*}
|
\begin{equation*}
\ \frac{2\pi}{3} \
\end{equation*}
|
\begin{equation*}
\frac{3\pi}{4}
\end{equation*}
|
\begin{equation*}
\frac{5\pi}{6}
\end{equation*}
|
\begin{equation*}
\ \ \ \pi \ \ \
\end{equation*}
|
\begin{equation*}
\frac{7\pi}{6}
\end{equation*}
|
\begin{equation*}
\frac{5\pi}{4}
\end{equation*}
|
\begin{equation*}
\ \frac{4\pi}{3} \
\end{equation*}
|
\begin{equation*}
\ \frac{3\pi}{2} \
\end{equation*}
|
|
\begin{equation*}
\cos(\theta)
\end{equation*}
|
\begin{equation*}
\ 1 \
\end{equation*}
|
\begin{equation*}
\frac{\sqrt{3}}{2}
\end{equation*}
|
\begin{equation*}
\frac{\sqrt{2}}{2}
\end{equation*}
|
\begin{equation*}
\frac{1}{2}
\end{equation*}
|
\begin{equation*}
0
\end{equation*}
|
\begin{equation*}
-\frac{1}{2}
\end{equation*}
|
\begin{equation*}
-\frac{\sqrt{2}}{2}
\end{equation*}
|
\begin{equation*}
-\frac{\sqrt{3}}{2}
\end{equation*}
|
\begin{equation*}
-1
\end{equation*}
|
\begin{equation*}
-\frac{\sqrt{3}}{2}
\end{equation*}
|
\begin{equation*}
-\frac{\sqrt{2}}{2}
\end{equation*}
|
\begin{equation*}
-\frac{1}{2}
\end{equation*}
|
\begin{equation*}
0
\end{equation*}
|
Using this table, we can sketch a graph of \(g(\theta)=\cos(\theta)\) as shown below. Plotting more points gives the full shape of the cosine function.
cosine graph
Like the sine function, we can track the value of the cosine function through the four quadrants of the unit circle as we sketch it on the graph. Since cosine corresponds to the \(x\) coordinates of points on the unit circle, the values of cosine are positive in quadrants 1 and 4 and negative in quadrants 2 and 3.
The applet below shows how the graphs of sine and cosine relate to the \(x\) and \(y\) coordinates of the point \(P\) on the unit circle. Move the slider to change the angle \(\theta\) and observe how the \(x\) and \(y\) values of point \(P\) relate to the outputs of \(\cos(\theta)\) and \(\sin(\theta)\text{.}\)
Exercises 1.3.4 Exercises
1. Identifying the Midline and Period.
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| A | B |
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| C | D |
(a) Which (if any) of the graphs A-D represents a graph of the function \(f(\theta)=\sin(\theta)\text{?}\)
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Select
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A
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B
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C
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D
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None of the graphs above
(b) Which (if any) of the graphs A-D represents a graph of the function \(g(\theta)=\cos(\theta)\)
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Select
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A
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B
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C
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D
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None of the graphs above
(c) What is the midline of the function \(y=\sin(\theta)\text{?}\)
The midline is \(y =\) .
(d) What is the midline of the function \(y=\cos(\theta)\text{?}\)
The midline is \(y =\) .
(e) What is the amplitude of the function \(\sin(\theta)\text{?}\)
The amplitude is .
(f) What is the amplitude of the function \(\cos(\theta)\text{?}\)
The amplitude is .
(g) What is the period of the function \(\sin(\theta)\text{?}\)
The period is .
(h) What is the period of the function \(\cos(\theta)\text{?}\)
The period is .
Hint.
Recall the definitions of midline, amplitude, and period.
2. Finding Values of Sine and Cosine.
This problem is similar to Problem 6 on your 1.3 Worksheet.
Suppose \(\displaystyle \sin( \theta ) = \frac{x}{7}\) and the angle \(\theta\) is in the second quadrant. Use the Pythagorean Identity to write an algebraic expression for \(\cos( \theta )\) in terms of \(x\text{.}\)
\(\cos( \theta )\) =
3. Finding Values of Sine and Cosine.
This problem is similar to Problem 6 on your 1.3 worksheet.
Suppose the angle \(\theta\) is in the first quadrant, \(0 \leq \theta \leq \pi /2\text{,}\) and \(\displaystyle \cos(\theta) = \frac{1}{6}\text{.}\) Find exact values (as fractions, not decimal approximations) for the following.
(a) \(\sin(\theta)\) =
(b) \(\tan(\theta)\) =
4. Applications of Sine and Cosine.
The top of a \(300\) foot vertical tower is to be anchored by cables that make an angle of \(60^{\circ}\) with the ground.
Please include units in your answers, and note that all trigonometric functions will be evaluated in degrees.
How long must each cable be?
How far from the base of the tower should anchors be placed?
5. Applications of Sine and Cosine.
A Ferris wheel with a diameter of 50 meters and a boarding platform at ground level is shown in the figure below. The wheel completes one full revolution every 6 minutes, and at time t = 0 an individual is at the 3:00 position and ascending.
(a) What is your initial height, at time t=0? . Please include in your answer.
(b) Which position will you be at after 90 seconds?
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What is your height after 90 seconds? Please include in your answer.
(c) Which position will you be at after 30 seconds?
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B
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C
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F
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What is the corresponding angle (from the 3:00 position) in degrees? degrees.
(d) What is your height after 30 seconds? . Please include in your answer.
(e) What is your height after 80 seconds? . Please include in your answer.
Hint.
a) Note that when \(t=0\text{,}\) an individual is at the 3:00 position.
b) What fractional part has an individual traveled after 90 seconds if a full revolution happens every 6 minutes? How can we use this to find the angle in relation to the 3:00 position?
c) What fractional part has an individual traveled after 30 seconds if a full revolution happens every 6 minutes? How can we use this to find the angle in relation to the 3:00 position?
d) Refer to your work done in the ladder problem of this assignment. Keep in mind that the length we find by doing this method is not the full height of an individual at this position.
e) How can we find the angle in relation to the 3:00 position? Refer to your work done in the ladder problem of this assignment to find the unknown length. Keep in mind that the length we find by doing this method is not the full height of an individual at this position.
6. Applications of Sine and Cosine.
This problem is similar to Problem 3 on your 1.3 worksheet.
Suppose you have a 14 foot ladder, and you place it against a wall so that the angle it forms with the ground is equal to \(45^{\circ}\text{.}\) Give your answers to the following rounded to two decimal places.
(a) What is the distance from the bottom of the ladder to the wall? feet.
(b) What is the distance from the top of the ladder to the ground? feet.
