Trigonometric Graphs - Amplitude and Periodicity

From the definition of the basic trigonometric functions as $x$- and $y$-coordinates of points on a unit circle, we see that by going around the circle one complete time $($or an angle of $2\pi),$ we return to the same point and therefore to the same $x$- and $y$-coordinates. This can be extended for going around the circle any multiple of times $($or any angle that is a multiple of $2\pi).$

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For example,

$0 = \sin (0) = \sin (0 + 2\pi) = \sin (0 + 2 \cdot 2\pi) = \cdots = \sin(0 + k \cdot 2\pi)$

for any integer $k$.

This shows the trigonometric functions are repeating. These functions are called periodic, and the period is the minimum interval it takes to capture an interval that when repeated over and over gives the complete function.

The periods of the basic trigonometric functions are as follows:

$\begin{array}{|c|c|} \hline \text{Function} & \text{Period}\\ \hline \sin (\theta) , \cos ( \theta ) & 2\pi\\ \hline \csc (\theta) , \sec ( \theta) & 2\pi \\ \hline \tan (\theta), \cot ( \theta ) & \pi\\ \hline \end{array}$

As we have seen, trigonometric functions follow an alternating pattern between hills and valleys. The amplitude of a trigonometric function is half the distance from the highest point of the curve to the bottom point of the curve:

$\text{(Amplitude)} = \frac{ \text{(Maximum) - (minimum)} }{2}.$

For example, if we consider the graph of $y=\sin(x)$

the amplitude is equal to $\frac{ \text{(Maximum) - (minimum)} }{2} = \frac{ 1 - (-1)}{2} = \frac{2}{2} = 1$. Similarly, the graph of $y=\cos(x)$ also has amplitude 1.

The amplitudes of the basic trigonometric functions are as follows:

$\begin{array}{|c|c|} \hline \text{Function} & \text{Amplitude}\\ \hline \sin (\theta) , \cos( \theta ) &1 \\ \hline \csc (\theta) , \sec ( \theta) & \text{N/A} \\ \hline \tan (\theta), \cot ( \theta ) & \text{N/A} \\ \hline \end{array}$

In transformation of trigonometric graphs, we see that multiplying a trigonometric function by a constant changes the amplitude. For example, the amplitude of the graph $y = 3 \sin(x)$ is $3$.

What are the amplitude and period of the graph $y = 5 \sin(x) - 2?$

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Since the trigonometric function $\sin(x)$ is multiplied by the constant $5$, the amplitude of the resulting graph is $5$. The result of the transformation is to shift the graph vertically by $-2$ in the $y$-direction and stretch the graph vertically by a factor of $5.$ Since the graph is not stretched horizontally, the period of the resulting graph is the same as the period of the function $\sin(x)$, or $2\pi$. $_\square$

What are the amplitude and period of the graph $y =-100\cos(1234\pi)?$

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It doesn't matter whether it's $-100$ or $100;$ at the turning points, $y = \pm100.$ So the amplitude is 100, and the fundamental period is $\frac{2\pi}{1234\pi} = \frac{1}{617}.\ _\square$

The fundamental period of a sine function $f$ that passes through the origin is given to be $3\pi$ and its amplitude is 5. Construct $f(x).$

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Since it passes through the origin, it must be of the form $f(x) = A \sin(kx)$ as $f(0) = 0$. Because its amplitude is 5, $f(x) = \pm 5 \sin(kx)$.

And because its fundamental period is $3\pi$, $\frac{2\pi}k = 3\pi \implies k = \frac23$. So $f(x) = \pm5 \sin\big( \frac23 x\big).\ _\square$

What are the fundamental period and amplitude of the function $f(x) =40\sin(2x) + 9\cos(2x)?$

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By $R$ method, we have $f(x) = \sqrt{40^2 + 9^2} \sin(2x + \alpha ) = 41 \sin(2x+ \alpha)$ for some constant $\alpha$ independent of $x$. So the amplitude is 41 while the fundamental period is $\frac {2\pi}2 = {\pi}$. $_\square$

What is the period of the function $h(x) = \sin\big( |123x|\big)?$

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A simple sketch of graph around the vicinity of $x = 0$ of $h(x) =\sin(123x) = \begin{cases} {\sin(123x) } \\ { -\sin(123x) } \end{cases}$ shows that it is not periodic. $_\square$

What are the amplitude and fundamental period of the function $f(x) = 64\cos^7(x) - 112\cos^5(x) + 56\cos^5(x) - 7\cos(x)?$

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By double angle formula and triple angle formula, we are able to obtain the fact that $f(x) = \cos(6x)$. Thus its amplitude is simply 1 and the fundamental period is $\frac 6 {2\pi} = \frac3{\pi}$. $_\square$

Note that we can also prove this using Chebyshev polynomials.