>Applications of the Derivative>How to Analyze the Cosine Function

How to Analyze the Cosine Function

Look at the function

f (x) = A \cos (c x + ϕ) + d .

Note! As $\sin x$ and $\cos x$ are so similar, you will be able to swap cosine for sine in the function and get approximately the same calculations as in the examples below.

You know that the graph of a normal cosine function is a wave, so it has multiple maxima, minima and zeros. The function $f (x)$ is very similar, but is shifted and stretched compared to the normal cosine function $\cos x$ .

Trigonometric function with zeros, maxima, minima and inflection points marked.

There are a number of very simple ways of finding the zeros, maxima, minima, and inflection points of the cosine function. We are going to look at them below.

Rule

Zeros

To find the zeros of

\cos x

, you set

x = \frac{π}{2} + n \cdot π

. To find the zeros for the more advanced cosine function

f (x) = A \cos (c x + ϕ) + d

you have to set $f (x) = 0$ and solve for $x$ .

Rule

Maxima

To find the maxima of

\cos x

, you set

\cos x

equal to 1. This means that any maximum will have a

y

-value of 1 and an

x

-value given by

x = 0 + n \cdot 2 π

. To find the maxima of the more advanced cosine function

f (x) = A \cos (c x + ϕ) + d

you use the following:

For $A > 0$ , the $y$ -value of a maximum is $A + d$ . If $A < 0$ , the $y$ -value of a maximum is $- A + d$ .
To find the corresponding $x$ -values, solve the equation $\cos (c x + ϕ) = 1$ for $x$ if $A > 0$ , and solve $\cos (c x + ϕ) = - 1$ if $A < 0$ .

Rule

Minima

To find the minima of

\cos x

, you set

\cos x

equal to

- 1

. This means that any minimum will have a

y

-value equal to

- 1

and an

x

-value given by

x = π + n \cdot 2 π

. To find the minima of

f (x) = A \cos (c x + ϕ) + d

you use the following:

If $A > 0$ , the $y$ -value of a minimum is $- A + d$ . If $A < 0$ , the $y$ -value of minimum is $A + d$ .
To find the corresponding $x$ -values, solve the equation $\cos (c x + ϕ) = - 1$ for $x$ if $A > 0$ , and solve $\cos (c x + ϕ) = 1$ if $A < 0$ .

Theory

Inflection Points

For

\cos x

, the inflection points are the same as the zeros. For

f (x) = A \cos (c x + ϕ) + d

the $y$ -value of an inflection point is $d$ . To find the $x$ -value, you solve $\cos (c x + ϕ) = 0$ .

Example 1

You are given the function
$f (x) = 4 \cos (π x + \frac{2 π}{3}) - 2$

Find the zeros, maxima, minima and inflection points of $f$ .

Zeros

Set $f (x) = 0$ and solve for $x$ : $\begin{array}{l} 4 \cos (π x + \frac{2 π}{3}) - 2 & = 0 \\ \cos (π x + \frac{2 π}{3}) & = \frac{1}{2} \end{array}$

The basic trigonometric equation has the solutions

\begin{array}{l} π x_{1} + \frac{2 π}{3} & = \frac{π}{3} + n \cdot 2 π \\ π x_{2} + \frac{2 π}{3} & = - \frac{π}{3} + n \cdot 2 π \end{array}

You solve these for $x$ and get

\begin{array}{l} π x_{1} & = - \frac{π}{3} + n \cdot 2 π \\ x_{1} & = - \frac{1}{3} + 2 n \\ π x_{2} & = - π + n \cdot 2 π \\ x_{2} & = - 1 + 2 n \end{array}

The zeros are thus

\dots, (- 1, 0), (- \frac{1}{3}, 0), (1, 0), (\frac{5}{3}, 0), \dots

Maxima

Since $4 > 0$ , the $y$ -coordinate of the maxima is

4 - 2 = 2

You find the $x$ -coordinate by solving the equation

\begin{array}{l} \cos (π x + \frac{2 π}{3}) & = 1 \\ π x + \frac{2 π}{3} & = 0 + n \cdot 2 π \\ π x & = 0 - \frac{2 π}{3} + n \cdot 2 π \\ π x & = - \frac{2 π}{3} + n \cdot 2 π \\ x & = - \frac{2}{3} + 2 n \end{array}

The maxima are thus

\begin{array}{l} \dots, & (- \frac{2}{3}, 2), (\frac{4}{3}, 2), (\frac{10}{3}, 2), \\ (\frac{16}{3}, 2), \dots \end{array}

\dots, (- \frac{2}{3}, 2), (\frac{4}{3}, 2), (\frac{10}{3}, 2), (\frac{16}{3}, 2), \dots

Note! As 0 and

- 0

is the same number, you only solve for one of the values—in this case 0. Thus, you end up with one equation—not the two you usually get when solving trigonometric equations with cosine in them.

Minima

Since $4 > 0$ , the $y$ -coordinate of the minima is

- 4 - 2 = - 6 .

You find the $x$ -coordinates by solving the equation

\begin{array}{l} \cos (π x + \frac{2 π}{3}) & = - 1 \\ π x + \frac{2 π}{3} & = π + n \cdot 2 π \\ π x & = - \frac{π}{3} + n \cdot 2 π \\ x & = - \frac{1}{3} + 2 n \end{array}

The minima are thus

\begin{array}{l} \dots, & (- \frac{1}{3}, - 6), (\frac{5}{3}, - 6), (\frac{11}{3}, - 6), \\ (\frac{17}{3}, - 6), \dots \end{array}

\begin{array}{l} \dots, (- \frac{1}{3}, - 6), (\frac{5}{3}, - 6), (\frac{11}{3}, - 6), (\frac{17}{3}, - 6), \dots \end{array}

Note! As

π

and

- π

give the same solution of the equation, you only solve for one of the values—in this case

π

. Thus, you end up with one equation—not two like you’re used to getting from trigonometric equations with cosine in them.

Inflection Points

You find the $y$ -value of the inflection points by reading off the value $d = - 2$ . You then find the $x$ -values by solving the equation $\cos (π x + \frac{2 π}{3}) = 0$ . The basic equation has the solutions $\begin{array}{l} π x_{1} + \frac{2 π}{3} & = \frac{π}{2} + n \cdot 2 π \\ π x_{2} + \frac{2 π}{3} & = - \frac{π}{2} + n \cdot 2 π \end{array}$

You solve these for $x$ and get:

\begin{array}{l} π x_{1} & = - \frac{2 π}{3} + \frac{π}{2} + n \cdot 2 π \\ = - \frac{π}{6} + n \cdot 2 π \\ x_{1} & = - \frac{1}{6} + 2 n \\ π x_{2} & = - \frac{2 π}{3} - \frac{π}{2} + n \cdot 2 π \\ = - \frac{7 π}{6} + n \cdot 2 π \\ x_{2} & = - \frac{7}{6} + 2 n \end{array}

The inflection points are thus

\begin{array}{l} \dots, & (- \frac{1}{6}, - 2), (\frac{5}{6}, - 2), (\frac{11}{6}, - 2), \\ (\frac{17}{6}, - 2), \dots \end{array}

\begin{array}{l} \dots, (- \frac{1}{6}, - 2), (\frac{5}{6}, - 2), (\frac{11}{6}, - 2), (\frac{17}{6}, - 2), \dots \end{array}

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