Symmetry in Trigonometric Graphs

The trigonometric functions cosine, sine, and tangent satisfy several properties of symmetry that are useful for understanding and evaluating these functions.

The cosine and sine functions satisfy the following properties of symmetry:

$$\begin{aligned} \cos(-\theta) &= \cos(\theta) \\ \sin(-\theta) &= -\sin(\theta). \end{aligned}$$

From the definition of cosine and sine in the unit circle,

$$x= \cos \theta \quad \text{ and } \quad y= \sin \theta.$$

We can see that for both $\theta$ and $-\theta$, $x$ remains the same. Thus, $\cos \theta=\cos (-\theta)$.

Similarly, we can see that the $y$ in two cases are additive inverse of each other. Thus, $\sin (-\theta)=-\sin\theta.\ _\square$

Now that we have the above identities, we can prove several other identities, as shown in the following example.

Prove the identities

$$\begin{aligned} \tan(-\theta) &= -\tan(\theta)\\ \cot(-\theta) &= -\cot(\theta)\\ \csc(-\theta) &= -\csc(\theta)\\ \sec(-\theta) &= \sec(\theta). \end{aligned}$$

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We have

$$\begin{aligned} \tan(-\theta) &=\frac{\sin(-\theta)}{\cos(-\theta)}=\frac{-\sin(\theta)}{\cos(\theta)}=-\tan(\theta)\\ \cot(-\theta) &=\frac{1}{\tan(-\theta)}=\frac{1}{-\tan(\theta)}=-\cot(\theta)\\ \csc(-\theta) &=\frac{1}{\sin(-\theta)}=\frac{1}{-\sin(\theta)}=-\csc(\theta)\\ \sec(-\theta) &=\frac{1}{\cos(-\theta)}=\frac{1}{\cos(\theta)}=\sec(\theta).\ _\square \end{aligned}$$

Using the properties of symmetry above, we can show that sine and cosine are special types of functions.

A function $f(x)$ is an even function if and only if for all real values of $x$, $f(-x)=f(x)$. In other words, the graph is symmetric about $y$-axis.

A function $f(x)$ is an odd function if and only if for all real values of $x$, $f(-x)=-f(x)$. In other words, the graph is symmetric about origin.
Also, $f(-0)=-f(0)\implies f(0)=0$. That is, an odd function must pass through the origin.

From this definition, the cosine function is an even function and the sine function is an odd function.

What symmetry is there between the angles $\theta$ and $(\theta + \pi)?$ If we plug in a few values for $\theta$, how do the basic trigonometric functions change?

By the properties of symmetry, we can write $\sin (\theta + \pi)$ in terms of $\sin(\theta)$ as follows:

$$\sin (\theta+\pi)=-\sin(\theta).$$

Similarly, we can write $\cos (\theta + \pi)$ in terms of $\cos(\theta)$ as follows:

$$\cos (\theta+\pi)=-\cos(\theta).$$

Determine whether the function

$$f(x) = \tan^2(x) + \cos(x)$$

is an odd function, an even function, or neither.

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The function satisfies

$$f(-x) = \tan^2(-x) + \cos(-x) = \tan^2(x) + \cos(x) = f(x)$$

since $\cos(x)$ is an even function. Therefore, $f(x)$ is an even function. $_\square$

Find a relationship between $\tan (\theta + \pi)$ and $\tan (\theta ).$

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Solution 1: We have

$$\begin{aligned} \tan (\theta + \pi) & = \frac{\sin(\theta + \pi ) }{\cos(\theta + \pi) } \\ &= \frac{-\sin(\theta)}{-\cos(\theta)}\\ &= \frac{\sin(\theta)}{\cos(\theta)}\\ &= \tan (\theta ). \end{aligned}$$

Therefore, $\tan (\theta + \pi) = \tan (\theta ).$

Solution 2: We have

$$\tan(x + \pi) = \frac{\tan(x) + \tan(\pi)}{ 1 - \tan(x) \tan(\pi)} = \frac{\tan(x) + 0}{ 1 - \tan(x) \cdot 0} = \tan(x).$$

This shows the period of the tangent function is at most $\pi$. $_\square$

Since the cosine function satisfies $\cos(-\theta) = \cos(\theta)$, the graph of the function $\cos(x)$ is symmetric about the $y$-axis. What is the symmetry satisfied by the graph of $\sin(x)?$

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Since the function $\sin(x)$ satisfies $\sin(-x) = \sin(x)$, the graph of $\sin(x)$ is symmetric about the origin. $_\square$

In general, for any even function $f(x)$, the the graph of $f(x)$ is symmetric about the $y$-axis; for any odd function $g(x)$, the graph of $g(x)$ is symmetric about the origin.

See Sine and Cosine graphs for more properties of the sine and cosine graphs.