Properties of Graphs of Trigonometric Functions

Trigonometric functions repeat every 2π radians. In this section we will discuss this and other properties of graphs, especially for the sinusoidal functions (sine and cosine).

First, recall that the domain of a function f(x) is the set of all numbers x for which the function is defined. For example, the domain of f(x) = sin x is the set of all real numbers, whereas the domain of f(x) = tan x is the set of all real numbers except x = ± \(\frac{π}{2}\) , ± \(\frac{3π}{2}\), ± \(\frac{5π}{2}\),.... The range of a function f(x) is the set of all values that f(x) can take over its domain. For example, the range of f(x) = sin x is the set of all real numbers between −1 and 1 (i.e. the interval [−1,1]), whereas the range of f(x) = tan x is the set of all real numbers, as we can see from their graphs.

A function f(x) is periodic if there exists a number p > 0 such that x+ p is in the domain of f(x) whenever x is, and if the following relation holds:

$$f (x+ p) = f (x) \; \text{for all} \; x \; \; \; \; (1)$$

There could be many numbers p that satisfy the above requirements. If there is a smallest such number p, then we call that number the period of the function f(x).

Example 2

What is the period of f(x) = sin 2x?

Solution: The graph of y = sin 2x is shown in Figure 5.2.1, along with the graph of y = sin x for comparison, over the interval [0,2π]. Note that sin 2x “goes twice as fast” as sin x.

For example, for x from 0 to \(\frac{π}{2}\) , sin x goes from 0 to 1, but sin 2x is able to go from 0 to 1 quicker, just over the interval [0, \(\frac{π}{4}\)]. While sin x takes a full 2π radians to go through an entire cycle (the largest part of the graph that does not repeat), sin 2x goes through an entire cycle in just π radians. So the period of sin 2x is π radians.

The above example made use of the graph of sin 2x, but the period can be found analytically. Since sin x has period 2π,1 we know that sin (x+2π) = sin x for all x. Since 2x is a number for all x, this means in particular that sin (2x+2π= sin 2x for all x. Now define f(x) = sin 2x. Then

$$f(x+π) = sin 2(x+π)$$ $$= sin (2x+2π)$$ $$= sin 2x \; \text{ as we showed above}$$ $$= f(x) $$

for all x, so the period p of sin 2x is at most π, by our definition of period. We have to show that p > 0 cannot be smaller than π. To do this, we will use a proof by contradiction. That is, assume that 0 < p < π, then show that this leads to some contradiction, and hence cannot be true. So suppose 0 < p < π. Then 0 < 2p < 2π, and hence

$$sin 2x = f(x)$$ $$= f(x+p) (\text{ since} \; p \text{ is the period of} \; f(x))$$ $$= sin 2(x + p)$$ $$= sin (2x + 2p)$$

for all x. Since any number u can be written as 2x for some x (i.e u = 2(u/2)), this means that sin u = sin (u+2p) for all real numbers u, and hence the period of sin x is as most 2p. This is a contradiction. Why? Because the period of sin x is 2π > 2p. Hence, the period p of sin 2x cannot be less than π, so the period must equal π.

The above may seem like a lot of work to prove something that was visually obvious from the graph (and intuitively obvious by the “twice as fast” idea). Luckily, we do not need to go through all that work for each function, since a similar argument works when sin 2x is replaced by sin ωx for any positive real number ω: instead of dividing 2π by 2 to get the period, divide by ω. And the argument works for the other trigonometric functions as well. Thus, we get:

If ω < 0, then use sin (−A) = −sin A and cos (−A) = cos A (e.g. sin (−3x) = −sin 3x).

Example 3

The period of y = cos 3x is \(\frac{2π}{3}\) and the period of y = cos \(\frac{1}{2}\)x is 4π. The graphs of both functions are shown in Figure 2:

We know that −1 ≤ sin x ≤ 1 and −1 ≤ cos x ≤ 1 for all x. Thus, for a constant A \(\ne\) 0,

−|A| ≤ A sin x ≤ |A| and −|A| ≤ A cos x ≤ |A|

for all x. In this case, we call |A| the amplitude of the functions y = A sin x and y = A cos x. In general, the amplitude of a periodic curve f(x) is half the difference of the largest and smallest values that f(x) can take:

$$\text{Amplitude of}\; f(x) = \frac{(\text{maximum of} f (x)) − (\text{minimum of} f (x))}{2}$$

In other words, the amplitude is the distance from either the top or bottom of the curve to the horizontal line that divides the curve in half, as in Figure 3.

Not all periodic curves have an amplitude. For example, tan x has neither a maximum nor a minimum, so its amplitude is undefined. Likewise, cot x, csc x, and sec x do not have an amplitude. Since the amplitude involves vertical distances, it has no effect on the period of a function, and vice versa.

Example 4

Find the amplitude and period of y = 3 cos 2x.

Solution: The amplitude is |3| = 3 and the period is \(\frac{2π}{2}\) = π. The graph is shown in Figure 4:

Example 5

Find the amplitude and period of y = 2−3 sin \(\frac{2π}{3}x\).

Solution: The amplitude of −3 sin \(\frac{2π}{3}x\) is |−3| = 3. Adding 2 to that function to get the function y = 2 − 3 sin \(\frac{2π}{3}x\) does not change the amplitude, even though it does change the maximum and minimum. It just shifts the entire graph upward by 2. So in this case, we have

$$\text{Amplitude} = \frac{\text{max} − \text{min}}{2} = \frac{5 - (-1)}{2} = \frac{6}{2} = 3$$

The period is \(\frac{2π}{\frac{2π}{3}}x\) = 3. The graph is shown in Figure 5:

Example 6

Find the amplitude and period of y = 2 sin (x²).

Solution: This is not a periodic function, since the angle that we are taking the sine of, x², is not a linear function of x, i.e. is not of the form ax+b for some constants a and b. Recall how we argued that sin 2x was “twice as fast” as sin x, so that its period was π instead of 2π. Can we say that sin (x²) is some constant times as fast as sin x? No. In fact, we see that the “speed” of the curve keeps increasing as x gets larger, since x² grows at a variable rate, not a constant rate. This can be seen in the graph of y = 2 sin (x²), shown in Figure 6.

Notice how the curve “speeds up” as x gets larger, making the “waves” narrower and narrower. Thus, y = 2 sin (x²) has no period. Despite this, it appears that the function does have an amplitude, namely 2. To see why, note that since |sin θ| ≤ 1 for all θ, we have

$$|2 sin (x^2)| = |2| • |sin (x^2)| ≤ 2 • 1 = 2 .$$

In the exercises you will be asked to find values of x such that 2 sin (x²) reaches the maximum value 2 and the minimum value −2. Thus, the amplitude is indeed 2.

Note: This curve is still sinusoidal despite not being periodic, since the general shape is still that of a “sine wave”, albeit one with variable cycles.

So far in our examples we have been able to determine the amplitudes of sinusoidal curves fairly easily. This will not always be the case.

Example 7

Find the amplitude and period of y = 3 sin x+4 cos x.

Solution: This is sometimes called a combination sinusoidal curve, since it is the sum of two such curves. The period is still simple to determine: since sin x and cos x each repeat every 2π radians, then so does the combination 3 sin x+4 cos x. Thus, y = 3 sin x+4 cos x has period 2π. We can see this in the graph, shown in Figure 7:

The graph suggests that the amplitude is 5, which may not be immediately obvious just by looking at how the function is defined. In fact, the definition y = 3 sin x+4 cos x may tempt you to think that the amplitude is 7, since the largest that 3 sin x could be is 3 and the largest that 4 cos x could be is 4, so that the largest their sum could be is 3+4 = 7. However, 3 sin x can never equal 3 for the same x that makes 4 cos x equal to 4 (why?).

There is a useful technique for showing that the amplitude of y = 3 sin x+4 cos x is 5. Let θ be the angle shown in the right triangle in Figure 5.2.8. Then cos θ = \(\frac{3}{5}\) and sin θ = \(\frac{4}{5}\) . We can use this as follows:

$$y = 3 sin x + 4 cos x$$ $$= 5(\frac{3}{5}sin x + \frac{4}{5}cos x)$$ $$= 5(cos θ sin x + sin θ cos x)$$ $$= 5 sin (x+θ) \; \text{ (by the sine addition formula)} $$

Thus, |y| = |5 sin (x+θ)| = |5| • |sin (x+θ)| ≤ (5)(1) = 5, so the amplitude of y = 3 sin x+4 cos x is 5.

In general, a combination of sines and cosines will have a period equal to the lowest common multiple of the periods of the sines and cosines being added. In Example 7, sin x and cos x each have period 2π, so the lowest common multiple (which is always an integer multiple) is 1 • 2π = 2π.

Example 8

Find the period of y = cos 6x+sin 4x.

Solution: The period of cos 6x is \(\frac{2π}{6}\) = \(\frac{π}{3}\), and the period of sin 4x is \(\frac{2π}{4}\) = \(\frac{π}{2}\). The lowest common multiple of \(\frac{π}{3}\) and \(\frac{π}{2}\) is π:

$$\frac{π}{3} = \frac{π}{3}$$	$$\frac{π}{2} = \frac{π}{2}$$
$$\frac{π}{3} = \frac{2π}{3}$$	$$\frac{π}{2} = π$$
$$\frac{π}{3} = π$$	$$\frac{π}{3} = π$$

Thus, the period of y = cos 6x+sin 4x is π. We can see this from its graph in Figure 9:

What about the amplitude? Unfortunately we cannot use the technique from Example 7, since we are not taking the cosine and sine of the same angle; we are taking the cosine of 6x but the sine of 4x. In this case, it appears from the graph that the maximum is close to 2 and the minimum is close to −2.

Generalizing Example 7, an expression of the form a sin ωx + b cos ωx is equivalent to \(\sqrt{a^2 +b^2}\) sin (x+θ), where θ is an angle such that cos θ = \(\frac{a}{\sqrt{a^2 +b^2}}\) and sin θ = \(\frac{b}{\sqrt{a^2 +b^2}}\). So y = a sin ωx + b cos ωx will have amplitude \(\sqrt{a^2 +b^2}\). Note that this method only works when the angle ωx is the same in both the sine and cosine terms. We have seen how adding a constant to a function shifts the entire graph vertically. We will now see how to shift the entire graph of a periodic curve horizontally.

Consider a function of the form y = A sin ωx, where A and ω are nonzero constants. For simplicity we will assume that A > 0 and ω > 0 (in general either one could be negative). Then the amplitude is A and the period is \(\frac{2π}{ω}\) . The graph is shown in Figure 10.

Now consider the function y = A sin (ωx−φ), where φ is some constant. The amplitude is still A, and the period is still \(\frac{2π}{ω}\) , since ωx−φ is a linear function of x. Also, we know that the sine function goes through an entire cycle when its angle goes from 0 to 2π. Here, we are taking the sine of the angle ωx−φ. So as ωx−φ goes from 0 to 2π, an entire cycle of the function y = A sin (ωx−φ) will be traced out. That cycle starts when

$$ωx−\phi = 0 ⇒ x = \frac{\phi}{ω}$$

and ends when

$$ωx−\phi = 2π ⇒ x = \frac{2π}{ω} + \frac{\phi}{ω}$$

Thus, the graph of y = A sin (ωx−φ) is just the graph of y = A sin ωx shifted horizontally by \(\frac{\phi}{ω}\), as in Figure 11. The graph is shifted to the right when φ > 0, and to the left when φ < 0. The amount \(\frac{\phi}{ω}\) of the shift is called the phase shift of the graph.

The phase shift is defined similarly for the other trigonometric functions.

Example 9

Find the amplitude, period, and phase shift of y = 3 cos (2x−π).

Solution: The amplitude is 3, the period is \(\frac{2π}{2}\) = π, and the phase shift is \(\frac{π}{2}\). The graph is shown in Figure 12:

Notice that the graph is the same as the graph of y = 3 cos 2x shifted to the right by \(\frac{π}{2}\), the amount of the phase shift.

Example 10

Find the amplitude, period, and phase shift of y = −2 sin \(\left(3x + \frac{π}{2} \right)\).

Solution: The amplitude is 2, the period is \(\frac{2π}{3}\), and the phase shift is \(\frac{-\frac{π}{2}}{3}\) = −\(\frac{π}{6}\) . Notice the negative sign in the phase shift, since 3x+π = 3x−(−π) is in the form ωx−φ. The graph is shown in Figure 13:

In engineering two periodic functions with the same period are said to be out of phase if their phase shifts differ. For example, sin (x− \(\frac{π}{6}\)) and sin x would be \(\frac{π}{6}\) radians (or 30º) out of phase, and sin x would be said to lag sin(x− \(\frac{π}{6}\)) by \(\frac{π}{6}\) radians, while sin(x− \(\frac{π}{6}\)) leads sin x by \(\frac{π}{6}\) radians. Periodic functions with the same period and the same phase shift are in phase.

The following is a summary of the properties of trigonometric graphs: