Graphs Of Cosine Sine And Tangent

Graph of f(x) = cosx

We will first begin to graph the function, f(x) = cosx by creating a table of values. For our extensive purposes, we will graph this function over the interval of [0, 2π], or simply, 1 rotation. We will use our values from the unit circle derivation for these points:

x	0	π/6	π/4	π/3	π/2	2π/3	3π/4	5π/6	π	7π/6	5π/4	4π/3	3π/2	5π/3	7π/4	11π/6	2π
f(x)	1	√3/2	√2/2	1/2	0	-1/2	-√2/2	-√3/2	-1	-√3/2	-√2/2	-1/2	0	1/2	√2/2	√3/2	1

When we graph this obtain:

Domain and Range of f(x) = sinx

The function f(x) = cosx extends from -∞ to ∞ in the x-direction. For example, we can let x = -100000 and evaluate f(-100000) to obtain a value. The y-direction of the function f(x) is restricted between -1 to 1 inclusive though.

(1)

\begin{align} -\infty < x < \infty \quad or \quad \quad x \in \mathbb{R} \end{align}

And:

(2)

\begin{align} -1 \leq f(x) \leq 1 \end{align}

Or in domain and range interval notation:

(3)

\begin{align} \mathbb{D}: (-\infty, \infty) \quad , \quad \mathbb{R}: [-1, 1] \end{align}

Graph of f(x) = sinx

We will graph f(x) = sinx in the same manner as we did f(x) = cosx. The table of values below are with regards to f(x) = sinx.

x	0	π/6	π/4	π/3	π/2	2π/3	3π/4	5π/6	π	7π/6	5π/4	4π/3	3π/2	5π/3	7π/4	11π/6	2π
f(x)	0	1/2	√2/2	√3/2	1	√3/2	√2/2	1/2	0	-1/2	-√2/2	-√3/2	-1	-√3/2	-√2/2	-1/2	0

The graph below is of f(x) = sinx:

Domain and Range of f(x) = sinx

The function f(x) = sinx extends from -∞ to ∞ in the x-direction as well. For example, we can let x = 345345 and evaluate f(345345) to obtain a value. The y-direction of the function f(x) is restricted between -1 to 1 inclusive though as well.

(4)

\begin{align} -\infty < x < \infty \quad or \quad x \in \mathbb{R} \end{align}

And:

(5)

\begin{align} -1 \leq f(x) \leq 1 \end{align}

Or in domain and range interval notation:

(6)

\begin{align} \mathbb{D}: (-\infty, \infty) \quad , \quad \mathbb{R}: [-1, 1] \end{align}

Graph of f(x) = tanx

We can clearly see that the functions f(x) = cosx and f(x) = sinx look rather similar. Unfortunately, this is not the case for f(x) = tanx. Let's first look at the table of values associated with f(x) = tanx:

x	0	π/6	π/4	π/3	π/2	2π/3	3π/4	5π/6	π	7π/6	5π/4	4π/3	3π/2	5π/3	7π/4	11π/6	2π
f(x)	0	1/√3	1	√3	d.n.e.	-√3	-1	-1/√3	0	1/√3	1	√3	d.n.e.	-√3	-1	-1/√3	0

Unfortunately, we have values of the function that do not exist (abbreviated d.n.e.). At these points, we have vertical asymptotes for the graph. In the case of f(x) = tanx, the equations of our vertical asymptotes are as follows:

(7)

\begin{align} x = \pi/2 \quad , \quad x = 3\pi/2 \end{align}

We know however that the domain of the function f(x) = tanx exists from -∞ to ∞, while the range of the functino f(x) exists from -1 to 1. Thus we can say that:

The graph below is of f(x) = tanx:

Domain and Range of f(x) = tanx

In the case of f(x) = tanx in the x-direction, we have some inconsistencies. From the table of values above, we know that x cannot equal π/2 , or 3π/2 as f(π/2) and f(3π/2) will result in an undefined result. In fact, x = π/2 , 3π/2 , 5π/2 , 7π/2 , etc… cannot be evaluated for f(x). We can write a general equation for these inconsistencies in the domain of tanx by writing a general rule for the asymptotes of our function

(8)

\begin{align} \quad x \neq \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, ... (2k - 1)\frac{\pi}{2}, \quad k \in \mathbb{I} \end{align}

For any value of k that exists as an integer {… -2, -1, 0, 1, 2, …} we will obtain an equation of a vertical asymptote for the graph f(x) = tanx. Thus we can say our domain of tanx to be:

(9)

\begin{align} \quad \mathbb{D}: x \in \mathbb{R} \quad , \quad x \neq (2k - 1)\frac{\pi}{2}, \quad k \in \mathbb{I} \end{align}

Additionally, as we can see from the graph of f(x) = tanx, the range of the function will extend from negative infinity to positive infinity. We can thus denote the range of tanx as:

(10)

\begin{align} \mathbb{R}: -\infty < y < \infty \quad or \quad \mathbb{R}: y \in \mathbb{R} \end{align}