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.

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.

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.

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]

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]

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.

1. 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.
2. 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.

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]

1. The graph is shown below.
   
2. The midline of the graph is [latex]D = 12.25[/latex] hours, the amplitude is 5.25 hours, and the period is 12 months.
3. On the longest day of the year, there are [latex]12.25 + 5.25 = 17.5[/latex] hours of daylight in Glasgow.
4. 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.

1. 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.
   
2. 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]