Algebra Refresher
Evaluate the function in part (a) and solve the equation in part (b).
1. [latex]f(x) = x^2 + 1[/latex]
- [latex]\displaystyle f(5)[/latex]
- [latex]\displaystyle f(x) = 5[/latex]
2. [latex]g(x) = x^3 - 2[/latex]
- [latex]\displaystyle g(-10)[/latex]
- [latex]\displaystyle g(x) = -10[/latex]
3. [latex]h(t) = \sqrt{2t + 6}[/latex]
- [latex]\displaystyle h(3)[/latex]
- [latex]\displaystyle h(t) = 3[/latex]
4. [latex]m(t) = 6\left(2^t\right)[/latex]
- [latex]\displaystyle m(3)[/latex]
- [latex]\displaystyle m(t) = 3[/latex]
[latex]\underline{\qquad\qquad\qquad\qquad}[/latex]
Algebra Refresher Answers
1.
- [latex]\displaystyle 26[/latex]
- [latex]\displaystyle \pm 2[/latex]
2.
- [latex]\displaystyle -102[/latex]
- [latex]\displaystyle -2[/latex]
3.
- [latex]\displaystyle 2\sqrt{3}[/latex]
- [latex]\displaystyle \dfrac{3}{2}[/latex]
4.
- [latex]\displaystyle 48[/latex]
- [latex]\displaystyle -1[/latex]
- Graph the trig functions of real numbers
- Solve trigonometric equations graphically
- Work with reference angles
- Solve trigonometric equations algebraically
- Evaluate trigonometric functions of real numbers
- Use trigonometric models #59–62
- Locate points on the graphs of the trigonometric functions
- Find the domain and range of a function
We can graph the circular functions [latex]y =\sin t,~ y = \cos t,[/latex] and [latex]y = \tan t ~[/latex] just as we graphed trigonometric functions of angles in degrees. The only difference is that we scale the horizontal axis in radians.
Consider the graph shown below.
You can probably recognize this graph as one cycle of [latex]y = \sin t{.}[/latex] For example, you can see that the graph completes one cycle at [latex]t = 2\pi[/latex] radians, or approximately 6.28. It reaches its maximum value, [latex]y = 1{,}[/latex] at [latex]t = \dfrac{\pi}{2}{,}[/latex] or approximately 1.57. You should also notice that [latex]y = 0[/latex] at [latex]t = \pi{,}[/latex] approximately 3.14.
Instead of scaling the horizontal axis with integers, we often use multiples of [latex]\pi{.}[/latex] Using such a scale, we can show the exact location of the intercepts of the graph and of its high and low points.
Graph one cycle of [latex]y = \sin t{,}[/latex] and scale the horizontal axis in multiples of [latex]\pi{.}[/latex]
Solution
We start by making a table of values for [latex]\sin t{,}[/latex] where [latex]t[/latex] is in radians. We’ll use the special values of [latex]t{,}[/latex] because we know the sines of those values.
| [latex]t[/latex] |
[latex]0[/latex] |
[latex]\dfrac{\pi}{6}[/latex] |
[latex]\dfrac{\pi}{4}[/latex] |
[latex]\dfrac{\pi}{3}[/latex] |
[latex]\dfrac{\pi}{2}[/latex] |
[latex]\dfrac{2\pi}{3}[/latex] |
[latex]\dfrac{3\pi}{4}[/latex] |
[latex]\dfrac{5\pi}{6}[/latex] |
[latex]\pi[/latex] |
| [latex]\sin t[/latex] |
[latex]0[/latex] |
[latex]0.500[/latex] |
[latex]0.707[/latex] |
[latex]0.866[/latex] |
[latex]1.000[/latex] |
[latex]0.866[/latex] |
[latex]0.707[/latex] |
[latex]0.500[/latex] |
[latex]0[/latex] |
In order to plot the points in the table, we scale the horizontal axis in multiples of [latex]\dfrac{\pi}{12}{.}[/latex] (Note that the special values are all multiples of [latex]\dfrac{\pi}{12}{.}[/latex] Thus, [latex]\dfrac{\pi}{6} = 2\left(\dfrac{\pi}{12}\right){,}[/latex] [latex]~ \dfrac{\pi}{4} = 3\left(\dfrac{\pi}{12}\right){,}[/latex] and so on.) The table lists values of [latex]\sin t[/latex] in the first two quadrants; in quadrants three and four, the sine values are negative.
| [latex]t[/latex] |
[latex]\pi[/latex] |
[latex]\dfrac{7\pi}{6}[/latex] |
[latex]\dfrac{5\pi}{4}[/latex] |
[latex]\dfrac{4\pi}{3}[/latex] |
[latex]\dfrac{3\pi}{2}[/latex] |
[latex]\dfrac{5\pi}{3}[/latex] |
[latex]\dfrac{7\pi}{4}[/latex] |
[latex]\dfrac{11\pi}{6}[/latex] |
[latex]2\pi[/latex] |
| [latex]\sin t[/latex] |
[latex]0[/latex] |
[latex]-0.500[/latex] |
[latex]-0.707[/latex] |
[latex]-0.866[/latex] |
[latex]-1.000[/latex] |
[latex]-0.866[/latex] |
[latex]-0.707[/latex] |
[latex]-0.500[/latex] |
[latex]0[/latex] |
The completed graph is shown below.
If we continued to plot points for values of [latex]t[/latex] greater than [latex]2 \pi[/latex] or less than 0, we would see the periodic behavior of the graph, as it repeats the same shape over each interval of length [latex]2 \pi{,}[/latex] just as it does for periods of [latex]360°[/latex] when the angles are measured in degrees.
Checkpoint 6.31.
- Complete the table of values for [latex]y = \cos t{.}[/latex]
| [latex]t[/latex] |
[latex]0[/latex] |
[latex]\dfrac{\pi}{6}[/latex] |
[latex]\dfrac{\pi}{4}[/latex] |
[latex]\dfrac{\pi}{3}[/latex] |
[latex]\dfrac{\pi}{2}[/latex] |
[latex]\dfrac{2\pi}{3}[/latex] |
[latex]\dfrac{3\pi}{4}[/latex] |
[latex]\dfrac{5\pi}{6}[/latex] |
[latex]\pi[/latex] |
| [latex]\cos t[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
| [latex]t[/latex] |
[latex]\pi[/latex] |
[latex]\dfrac{7\pi}{6}[/latex] |
[latex]\dfrac{5\pi}{4}[/latex] |
[latex]\dfrac{4\pi}{3}[/latex] |
[latex]\dfrac{3\pi}{2}[/latex] |
[latex]\dfrac{5\pi}{3}[/latex] |
[latex]\dfrac{7\pi}{4}[/latex] |
[latex]\dfrac{11\pi}{6}[/latex] |
[latex]2\pi[/latex] |
| [latex]\cos t[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
[latex]\hphantom{0000}[/latex] |
- Graph one cycle of [latex]y = \cos t{,}[/latex] and scale the horizontal axis in multiples of [latex]\pi{.}[/latex] Use the grid below.
Solution
-
| [latex]t[/latex] |
[latex]0[/latex] |
[latex]\dfrac{\pi}{6}[/latex] |
[latex]\dfrac{\pi}{4}[/latex] |
[latex]\dfrac{\pi}{3}[/latex] |
[latex]\dfrac{\pi}{2}[/latex] |
[latex]\dfrac{2\pi}{3}[/latex] |
[latex]\dfrac{3\pi}{4}[/latex] |
[latex]\dfrac{5\pi}{6}[/latex] |
[latex]\pi[/latex] |
| [latex]\cos t[/latex] |
[latex]1[/latex] |
[latex]\dfrac{\sqrt{3}}{2}[/latex] |
[latex]\dfrac{1}{\sqrt{2}}[/latex] |
[latex]\dfrac{1}{2}[/latex] |
[latex]0[/latex] |
[latex]\dfrac{-1}{2}[/latex] |
[latex]\dfrac{-1}{\sqrt{2}}[/latex] |
[latex]\dfrac{-\sqrt{3}}{2}[/latex] |
[latex]-1[/latex] |
-
| [latex]t[/latex] |
[latex]\pi[/latex] |
[latex]\dfrac{7\pi}{6}[/latex] |
[latex]\dfrac{5\pi}{4}[/latex] |
[latex]\dfrac{4\pi}{3}[/latex] |
[latex]\dfrac{3\pi}{2}[/latex] |
[latex]\dfrac{5\pi}{3}[/latex] |
[latex]\dfrac{7\pi}{4}[/latex] |
[latex]\dfrac{11\pi}{6}[/latex] |
[latex]2\pi[/latex] |
| [latex]\cos t[/latex] |
[latex]-1[/latex] |
[latex]-\dfrac{\sqrt{3}}{2}[/latex] |
[latex]-\dfrac{1}{\sqrt{2}}[/latex] |
[latex]-\dfrac{1}{2}[/latex] |
[latex]0[/latex] |
[latex]\dfrac{1}{2}[/latex] |
[latex]\dfrac{1}{\sqrt{2}}[/latex] |
[latex]\dfrac{\sqrt{3}}{2}[/latex] |
[latex]1[/latex] |
Finally, we consider the graph of the tangent function. Once again, the only difference between this new graph and our old version of the tangent graph in degrees is that the horizontal axis is scaled in radians.
Graph the tangent function, [latex]y = \tan t{,}[/latex] where [latex]t[/latex] is measured in radians.
Solution
We begin by making a table of values for the tangent function. We choose the special values for [latex]t[/latex] between [latex]0[/latex] and [latex]\pi{.}[/latex]
| [latex]t[/latex] |
[latex]0[/latex] |
[latex]\dfrac{\pi}{6}[/latex] |
[latex]\dfrac{\pi}{4}[/latex] |
[latex]\dfrac{\pi}{3}[/latex] |
[latex]\dfrac{\pi}{2}[/latex] |
[latex]\dfrac{2\pi}{3}[/latex] |
[latex]\dfrac{3\pi}{4}[/latex] |
[latex]\dfrac{5\pi}{6}[/latex] |
[latex]\pi[/latex] |
| [latex]\tan t[/latex] |
[latex]0[/latex] |
[latex]0.577[/latex] |
[latex]1[/latex] |
[latex]1.732[/latex] |
[latex]---[/latex] |
[latex]-1.732[/latex] |
[latex]-1[/latex] |
[latex]-0.577[/latex] |
[latex]0[/latex] |
Plot the points and connect them with smooth curves, remembering that [latex]\tan (\dfrac{\pi}{2})[/latex] is undefined. The graph repeats for values of [latex]t[/latex] between [latex]\pi[/latex] and [latex]2\pi{.}[/latex] So while the graph in degrees had vertical asymptotes at [latex]\theta = 90°[/latex] and [latex]\theta = 270°{,}[/latex] this graph has vertical asymptotes at [latex]t = \dfrac{\pi}{2}[/latex] and [latex]t = \dfrac{3\pi}{2}{.}[/latex] The figure at right shows the graph of [latex]y = \tan t[/latex] for [latex]0 \le t \le 2\pi{.}[/latex]
Checkpoint 6.34.
Sketch a graph of the tangent function [latex]y =\tan t[/latex] and scale the horizontal axis in integers, as shown below. Label the [latex]t[/latex]-intercepts and the vertical asymptotes with their coordinates.
Solution
We can use the graphs of the circular functions to find solutions to trigonometric equations.
Use a graph to find all solutions of [latex]\cos t = -0.62[/latex] between [latex]0[/latex] and [latex]2\pi{.}[/latex]
Solution
The figure shows a graph of [latex]y = \cos t{,}[/latex] with [latex]t[/latex] in radians, and the horizontal line [latex]y = -0.62{.}[/latex] The solutions of the equation are the [latex]t[/latex]-coordinates of the intersection points, and these appear to be about [latex]t = 2.25[/latex] and slightly over [latex]4.0{,}[/latex] perhaps [latex]4.05{.}[/latex] You can check that these values roughly satisfy the equation; it is difficult to read the graph with any greater accuracy.
Checkpoint 6.36.
Use a graph to find the two angles between [latex]0[/latex] and [latex]2\pi[/latex] that satisfy [latex]\sin t = -0.85{.}[/latex]
Solution
Approximately 4.14 and 5.25
We can also solve equations algebraically, using a calculator or computer to obtain more accurate values for the solutions.
- Solve the equation [latex]~~\cos t = -0.62~~[/latex] algebraically, for [latex]0 \le t \le 2\pi{.}[/latex]
- Find all solutions of the equation.
Solution
- The cosine is negative in the second and third quadrants, so we expect to find our answers between [latex]t = 1.57[/latex] and [latex]t = 4.71{.}[/latex] We use a calculator to find one of the angles: [latex]\cos^{-1}~ (-0.62) = 2.23953903[/latex] Rounded to hundredths, the second quadrant angle is [latex]t_1 = 2.24[/latex] radians. To find the other angle, we first find the reference angle for [latex]t_1{:}[/latex]
reference angle [latex]= \pi - t_1 = 0.90[/latex]The third quadrant angle with the same reference angle is[latex]t_2 = \pi + 0.90 = 4.04[/latex]Thus, the two solutions are [latex]t_1 = 2.24[/latex] and [latex]t_2 = 4.04[/latex] radians, as shown in the figure.
- All angles coterminal with [latex]t_1[/latex] and [latex]t_2[/latex] are also solutions of the equation [latex]~~\cos t = -0.62~~{.}[/latex] So adding or subtracting multiples of [latex]2 \pi[/latex] to [latex]t_1[/latex] and [latex]t_2[/latex] produces all the solutions. For example, [latex]2.24 + 2\pi = 8.52[/latex] and [latex]4.04 - 2 \pi = -2.24[/latex] are solutions. We write the solutions as [latex]t = 2.24 \pm 2 \pi k[/latex] and [latex]t = 4.04 \pm 2 \pi k{,}[/latex] where [latex]k[/latex] is an integer.
Checkpoint 6.38.
- Solve the equation [latex]~~\sin t = -0.85~~[/latex] algebraically, for [latex]0 \le t \le 2\pi{.}[/latex]
- Find all solutions of the equation.
Solution
- [latex]4.16[/latex] and [latex]5.27[/latex]
- [latex]4.16 \pm 2\pi k[/latex] and [latex]5.27 \pm 2\pi k[/latex], where [latex]k[/latex] is an integer.
We can find exact values for the solutions of equations involving the special values without using a calculator.
Find exact values for all the solutions of [latex]~~\tan t = 1{.}[/latex]
Solution
Because [latex]\tan \dfrac{\pi}{4} = 1{,}[/latex] one of the solutions is [latex]t_1 = \dfrac{\pi}{4}{.}[/latex] Remember that there are two solutions between [latex]0[/latex] and [latex]2\pi{;}[/latex] and the tangent function is positive in the third quadrant. The third quadrant angle with reference angle [latex]\dfrac{\pi}{4}[/latex] is [latex]~~\pi + \dfrac{\pi}{4} = \dfrac{5\pi}{4},~~[/latex] so the second solution is [latex]t_2 = \dfrac{5\pi}{4}{.}[/latex]
All other solutions of the equation [latex]\tan t = 1[/latex] can be found by adding (or subtracting) multiples of [latex]2\pi[/latex] to these two solutions. We write the solutions as
[latex]t = \dfrac{\pi}{4} \pm 2\pi k,~~ \dfrac{5\pi}{4} \pm 2\pi k,~~ where [latex]k[/latex] is an integer.
[/latex]
Checkpoint 6.40.
Find exact values for all the solutions of [latex]~~\tan t = -\sqrt{3}{.}[/latex]
Solution
[latex]\dfrac{2\pi}{3} \pm 2\pi k{,}[/latex] [latex]~ \dfrac{5\pi}{3} \pm 2\pi k[/latex], where [latex]k[/latex] is an integer.
Now that we have defined trigonometric functions of real numbers, we can describe periodic phenomena as functions of time (or other variables besides angles).
For example, we began this chapter with a Ferris wheel of radius 100 feet that rotates once every 8 minutes. If you board the Ferris wheel at the bottom, your height is given as a function of time by
[latex]h = f(t) = 100 - 100\cos \left(\dfrac{\pi}{4}t\right)[/latex]
where [latex]t[/latex] is measured in minutes after boarding.
Thus, after [latex]{2}[/latex] minutes your height is
[latex]f({2}) = 100 - 100\cos \left(\dfrac{\pi}{2}\right)\\ = 100 - 100(0) = 100[/latex]
and after [latex]{4}[/latex] minutes your height is
[latex]f({4}) = 100 - 100\cos (\pi)[/latex]
The graph of [latex]h = f(t)[/latex] is shown above. From the graph, we see that the midline of the function is [latex]h = 100{,}[/latex] its amplitude is 100, and its period is 8 (which is reasonable because the Ferris wheel rotates every 8 minutes). You can review period, midline, and amplitude of periodic functions in Section 4.3.
The formula[latex]D(t) = 12.25 - 5.25\cos \left(\dfrac{\pi}{6}t\right)[/latex]gives the number of hours of daylight in Glasgow, Scotland, as a function of time, in months after January 1.
- Graph the function on your calculator (make sure the calculator is set in radian mode).
- State the midline, amplitude, and period of the graph.
- How many hours of daylight does Glasgow enjoy on the longest day of the year?
- In which months does Glasgow experience less than 8 hours of light per day?
Solution
- The graph is shown below.
- The midline of the graph is [latex]D = 12.25[/latex] hours, the amplitude is 5.25 hours, and the period is 12 months.
- On the longest day of the year, there are [latex]12.25 + 5.25 = 17.5[/latex] hours of daylight in Glasgow.
- From the graph, we see that [latex]D(t) \lt 8[/latex] for [latex]t[/latex] between 0 and 2, or in January and February. [latex]D(t)[/latex] is also less than 8 for a short time between [latex]t = 11[/latex] and [latex]t = 12{,}[/latex] at the end of December.
Checkpoint 6.42.
The pistons in an automobile engine move up and down in the cylinders. If [latex]t[/latex] is in milliseconds, the distance from the top of the piston to the top of the cylinder is given in centimeters by
[latex]D = 7 + 6\sin (4\pi t)[/latex]
- Graph the function on your calculator (make sure the calculator is set in radian mode).
- State the midline, amplitude, and period of the graph.
- Find the largest and the smallest clearance between the piston and the top of the cylinder.
Solution
- Midline: [latex]D = 7{,}[/latex] amplitude: [latex]6{,}[/latex] period: [latex]0.5[/latex] millisec
- Largest: 13 cm, smallest: 1 cm
The domain of a function is the set of all possible input values. For many familiar functions, the domain is the set of all real numbers. In particular, the domain of any linear or quadratic function is the set of all real numbers.
Consider the functions [latex]f(x) = 2x + 5[/latex] and [latex]g(x) = x^2 - 4[/latex] shown below. We can use any real number as an input for either of these functions, and their graphs extend across the entire [latex]x[/latex]-axis.
However, for other types of functions we must sometimes exclude certain values from the domain. For example, the domain of the function [latex]h(x) = \sqrt{x + 3}[/latex] is restricted because we cannot take the square root of a negative number. We must have [latex]x + 3 \ge 0{,}[/latex] so the domain of the function consists of all [latex]x \ge -3{.}[/latex]
You can see the domain of a function in its graph; notice that there are no points on the graph of [latex]h(x) = \sqrt{x + 3}[/latex] with [latex]x[/latex]-coordinates less than [latex]-3{.}[/latex]
The range of a function is the set of all output values for the function. We can also see the range of a function in its graph; it is the set of all [latex]y[/latex]-values for points on the graph.
However, the range of [latex]g(x) = x^2 - 4[/latex] consists of all [latex]y \ge -4{,}[/latex] because there are no points on the graph with [latex]y[/latex]-coordinate less than [latex]-4{.}[/latex]
The range of [latex]h(x) = \sqrt{x + 3}[/latex] consists of all [latex]y \ge 0{.}[/latex]
The domain of a function is the set of all possible input values.
The range of a function is the set of all output values for the function.
Find the domain and range of each function.
-
[latex]\displaystyle f(x)= 1-x^2, g(x) = \dfrac{1}{x - 3}[/latex]
Solution
- There are no restrictions on the input values for this function, so its domain is all real numbers. The graph is a parabola with high point at [latex](0,1){,}[/latex] so the range of the function is all [latex]y \le 1{.}[/latex] See the figure below.
- Because we cannot divide by zero, we cannot allow [latex]x = 3[/latex] for this function. Thus, the domain of [latex]g[/latex] is all real numbers except [latex]3{.}[/latex] The graph has a horizontal asymptote at [latex]y = 0{,}[/latex] and you can see that there is no input that will produce an output of zero. So the range of [latex]g[/latex] is all real numbers except [latex]0{.}[/latex]
Checkpoint 6.44.
Use the formula and a graph to find the domain and range.
- [latex]\displaystyle f(x) = \dfrac{2x}{x + 4}[/latex]
- [latex]\displaystyle g(x) = 1 + \sqrt{5 - x}[/latex]
Solution
- Dom ([latex]f[/latex]): all real numbers except [latex]-4{,}[/latex] Rge ([latex]f[/latex]): all real numbers except [latex]2[/latex]
- Dom ([latex]g[/latex]): [latex](-\infty, 5]{,}[/latex] Rge ([latex]g[/latex]): [latex][1, \infty)[/latex]
What about the domain and range of the trigonometric functions? The sine and cosine both include all real numbers in their domains; we can find the sine or cosine of any number. Because the output values of the sine and cosine are both defined by the coordinates of points on a unit circle, their values cannot be greater than [latex]1[/latex] or less than [latex]-1{.}[/latex] The range of both sine and cosine is the interval [latex][-1,1]{.}[/latex]
The tangent function is undefined at odd multiples of [latex]\dfrac{\pi}{2}[/latex] (that is, at [latex]\cdots,~ \dfrac{-3\pi}{2},~ \dfrac{-\pi}{2},~ \dfrac{\pi}{2},~ \dfrac{3\pi}{2},~\cdots[/latex]), so those values must be excluded from its domain. However, the output values of the tangent function increase without bound as the input approaches [latex]\dfrac{\pi}{2}[/latex] from the left and decrease from the right. The range of the tangent is all real numbers.
These facts about the three trigonometric functions appear in the Section 6.3 Summary.
Concepts
-
- We can use circular functions of real numbers to describe periodic phenomena.
- The domain of a function is the set of all possible input values. The range of a function is the set of all output values for the function.
- We can use a graph to solve trigonometric equations, or the inverse trig keys on a calculator or computer. We can find exact values for the solutions of equations involving the special values without using a calculator.
- [latex]f(x) = \sin x[/latex].
Domain: all real numbers
Range: [latex][-1,1][/latex]
Period: [latex]2\pi[/latex]
- [latex]g(x) = \cos x[/latex].
Domain: all real numbers
Range: [latex][-1,1][/latex]
Period: [latex]2\pi[/latex]
- [latex]h(x) = \tan x[/latex].
Domain: all real numbers except [latex]\cdots,~ \dfrac{-3\pi}{2},~ \dfrac{-\pi}{2},~ \dfrac{\pi}{2},~ \dfrac{3\pi}{2},~\cdots[/latex]
Range: all real numbers
Period: [latex]\pi[/latex]
Study Questions
- How do the graphs of the circular functions differ from the graphs of the trigonometric functions of angles in degrees?
- Use guide points to sketch graphs of [latex]y = \cos t[/latex] and [latex]y = \tan t{.}[/latex] For each equation below, suppose that [latex]\omega[/latex] is one of the solutions between [latex]0[/latex] and [latex]2\pi{.}[/latex] Use the diagram to find the other solution.
- [latex]\displaystyle \sin t = k[/latex]
- [latex]\displaystyle \cos t = k[/latex]
- [latex]\displaystyle \tan t = k[/latex]