Apply transformations to trigonometric functions

Transformed sine, cosine, and tangent functions are represented in forms such as $y=a\sin(b(x-h))+k$ , with $a$ determining vertical stretch and reflection, $b$ determining horizontal scale, $h$ the phase shift, and $k$ the vertical translation. These parameters are connected to graphs, key points, periods, domain, range, and tangent asymptotes, including periods $2\pi/|b|$ for sine and cosine and $\pi/|b|$ for tangent; the scope is standard real-valued transformations, not inverse, parametric, or more advanced generalized-function analysis.

Detailed Explanation: Apply transformations to trigonometric functions

A transformed trigonometric function can be written as

y=a\sin(b(x-h))+k,\qquad y=a\cos(b(x-h))+k,

y=a\tan(b(x-h))+k.

The parameters have these meanings:

$a$ : vertical stretch or compression; if $a<0$ , reflect across the $x$ -axis.
$b$ $b$ : horizontal scale factor. The period is
- $\dfrac{2\pi}{ \vert b \vert }$ for sine and cosine,
- $\dfrac{\pi}{ \vert b \vert }$ for tangent.
$h$ : horizontal shift, right if $h>0$ and left if $h<0$ .
$k$ : vertical shift, up if $k>0$ and down if $k<0$ .

Example

Analyze and sketch

y=-2\cos\left(3\left(x-\frac{\pi}{6}\right)\right)+1.

Step 1: Identify the parameters

Compare the equation with

y=a\cos(b(x-h))+k.

Therefore,

a=-2,\qquad b=3,\qquad h=\frac{\pi}{6},\qquad k=1.

This tells us:

The graph is reflected across the $x$ -axis because $a$ is negative.
It has a vertical stretch by a factor of $2$ .
It is horizontally compressed because $b=3$ .
It shifts right by $\dfrac{\pi}{6}$ .
It shifts up by $1$ .

Step 2: Find the period

For cosine,

\text{Period}=\frac{2\pi}{ \vert b \vert }.

So,

\text{Period}=\frac{2\pi}{3}.

A cosine cycle is divided into four equal sections, so the distance between key points is

\frac{\text{Period}}{4} =\frac{2\pi/3}{4} =\frac{\pi}{6}.

Step 3: Find the key points

The first key point occurs at the phase shift:

x=h=\frac{\pi}{6}.

Starting with the standard cosine values, use the five points across one cycle:

1,\quad 0,\quad -1,\quad 0,\quad 1.

Substitute these into the transformed function $y=-2\cos(\cdots)+1$ :

\begin{array}{c|c|c} x & \cos(\cdots) & y=-2\cos(\cdots)+1\\ \hline \frac{\pi}{6} & 1 & -1\\ \frac{\pi}{3} & 0 & 1\\ \frac{\pi}{2} & -1 & 3\\ \frac{2\pi}{3} & 0 & 1\\ \frac{5\pi}{6} & 1 & -1 \end{array}

The key points are

\left(\frac{\pi}{6},-1\right),\ \left(\frac{\pi}{3},1\right),\ \left(\frac{\pi}{2},3\right),\ \left(\frac{2\pi}{3},1\right),\ \left(\frac{5\pi}{6},-1\right).

Plot these points and connect them with a smooth cosine curve. The pattern repeats every $\dfrac{2\pi}{3}$ .

Step 4: State the range and domain

The midline is

y=k=1.

The amplitude is

\vert a \vert =2.

Therefore, the maximum value is

1+2=3,

and the minimum value is

1-2=-1.

So the range is

-1\le y\le 3.

The domain of a sine or cosine function is all real numbers:

x\in\mathbb{R}.

Thus, the transformed graph has period $\dfrac{2\pi}{3}$ , range $[-1,3]$ , and domain $\mathbb{R}$ .

Learn by doing: Apply transformations to trigonometric functions

Click a topic below to practice the foundational skills you'll need, learn the steps, or master this skill

Practice with unlimited practice problems

Sinusoidal Function Parameters (4 Params) - Function to Graph

Earned ?