What is the amplitude and vertical shift of the function $y = -2 \sin(x) - 1$?

The amplitude is $|-2| = 2$. The vertical shift is down 1 unit ($c = -1$).

What is the period of the function $y = 3 \cos(2x + \frac{\pi}{3}) - 1$?

The period is $\pi$. It is calculated as $\frac{2\pi}{k}$ where $k = 2$.

How do you determine the phase shift of a sine or cosine graph?

To find the phase shift, the function must be in the form $y = a \sin(k(x - d)) + c$. Factor out the $k$ value first. Then, $d$ is the phase shift. If the term is $(x - d)$, the shift is to the right. If the term is $(x + d)$, the shift is to the left.

How does a negative amplitude affect the graph of a sine or cosine function?

A negative 'a' value does not change the amplitude, which is always positive ($|a|$). However, the negative sign causes the graph to reflect across the horizontal midline of the wave.

How to Graph Sine and Cosine Functions | Examples & Transformations

Learn how to sketch the sine and cosine functions of the general forms:

$$ y = a \sin\bigl(k (x - d)\bigr) + c \quad \text{and} \quad y = a \cos\bigl(k (x - d)\bigr) + c $$

With clear explanations, step-by-step examples, and detailed solutions, this introduction covers key concepts such as amplitude, period, phase shift, and vertical translation to help you master graphing trigonometric functions.

Graphing Parameters

Amplitude: $ |a| $

Period: $ \dfrac{2\pi}{|k|} $

Horizontal Shift (Phase Shift): $d$. To the left if $-d$ is positive, and to the right if $-d$ is negative.

Vertical Shift: $c$. Up if $c$ is positive, and down if $c$ is negative.

Unit Circle Foundations

In order to sketch transformed sine and cosine functions, we need to know how to sketch basic sine and cosine functions. The unit circle (radius = 1) gives the values of $\sin(x)$ and $\cos(x)$ at 5 key points which can be used to graph more complex functions. The coordinates of any point on the unit circle give the cosine and sine of the angle in standard position corresponding to that point.

The angle $x = 0$ corresponds to the point: $ (1, 0) = (\cos(0), \sin(0)) $
The angle $x = \pi/2$ (or 90°) corresponds to the point: $ (0, 1) = (\cos(\pi/2), \sin(\pi/2)) $
The angle $x = \pi$ (or 180°) corresponds to the point: $ (-1, 0) = (\cos(\pi), \sin(\pi)) $
The angle $x = 3\pi/2$ (or 270°) corresponds to the point: $ (0, -1) = (\cos(3\pi/2), \sin(3\pi/2)) $
The angle $x = 2\pi$ (or 360°) corresponds to the point: $ (1, 0) = (\cos(2\pi), \sin(2\pi)) $

Sketching Sine and Cosine Functions: Examples with Detailed Solutions

Example 1: Vertical Stretch and Shift

Sketch the graph of $y = 2 \cos(x) + 1$ over one period.

Solution:

Graphing Parameters

Amplitude: $|2| = 2$
Period: $2\pi$
Vertical Shift: $1$ (up 1 unit)
Horizontal Shift: $0$

Three steps to graph the function $y = 2 \cos(x) + 1$:

1) We start by sketching $y = \cos(x)$ using the values of $x$ and $y$ from the unit circle (blue graph below).

$$ \begin{array}{|c|c|c|c|c|c|} \hline x & 0 & \dfrac{\pi}{2} & \pi & \dfrac{3\pi}{2} & 2\pi \\ \hline y & 1 & 0 & -1 & 0 & 1 \\ \hline \end{array} $$

2) We then sketch $y = 2 \cos(x)$ by stretching $y = \cos(x)$ by 2 (green graph below).

3) Finally, sketch $y = 2 \cos(x) + 1$ by shifting up 1 unit (red graph below).

Example 2: Vertical Stretch, Shift, and Reflection

Sketch the graph of $y = - 2 \sin(x) - 1$ over one period.

Solution:

Graphing Parameters

Amplitude: $|-2| = 2$
Period: $2\pi$
Vertical Shift: $-1$ (down 1 unit)
Horizontal Shift: $0$

Three steps to graph the function $y = - 2 \sin(x) - 1$:

1) We start by sketching $y = \sin(x)$ using the values of $x$ and $y$ from the unit circle (blue graph below).

$$ \begin{array}{|c|c|c|c|c|c|} \hline x & 0 & \dfrac{\pi}{2} & \pi & \dfrac{3\pi}{2} & 2\pi \\ \hline y & 0 & 1 & 0 & -1 & 0 \\ \hline \end{array} $$

2) We then sketch $y = - 2 \sin(x)$ by stretching $y = \sin(x)$ by 2 and reflecting it on the x-axis (green graph below).

3) Finally, sketch $y = - 2 \sin(x) - 1$ by shifting down 1 unit (red graph below).

Example 3: Period Change and Phase Shift

Sketch the graph of $y = 3 \cos(2x + \dfrac{\pi}{3}) - 1$ over one period.

Solution:

Graphing Parameters

Amplitude: $|3| = 3$
Period: $\dfrac{2\pi}{2} = \pi$
Vertical Shift: $-1$ (down 1 unit)

Horizontal Shift: Because of the term $\dfrac{\pi}{3}$, the graph is shifted horizontally. We first rewrite the given function as:
$$ y = 3 \cos\left[ 2\left( x + \dfrac{\pi}{6} \right) \right] - 1 $$
and we can now write the shift as being equal to $\dfrac{\pi}{6}$ to the left.

Three steps to graph the function $y = 3 \cos(2x + \dfrac{\pi}{3}) - 1$:

1) We start by sketching $3 \cos(2x)$ with minimum and maximum values -3 and +3 over one period $= \pi$ (blue graph below).

2) We then sketch $y = 3 \cos(2x) - 1$ translating the previous graph 1 unit down (green graph below).

3) We now shift the previous graph $\dfrac{\pi}{6}$ to the left (red graph below) so that the sketched period starts at $-\dfrac{\pi}{6}$ and ends at $-\dfrac{\pi}{6} + \pi = \dfrac{5\pi}{6}$ which is one period $= \pi$.

Example 4: Decimal Parameters

Sketch the graph of $y = -0.2 \sin\left(0.5x - \dfrac{\pi}{6} \right) + 0.1$ over one period.

Solution:

Graphing Parameters

Amplitude: $|-0.2| = 0.2$
Period: $\dfrac{2\pi}{0.5} = 4\pi$
Vertical Shift: $0.1$ (up 0.1 unit)

Horizontal Shift: Because of the term $- \dfrac{\pi}{6}$, the graph is shifted horizontally. We first rewrite the given function as:
$$ y = -0.2 \sin\left( 0.5\left(x - \dfrac{\pi}{3}\right) \right) + 0.1 $$
and we can now write the shift as being equal to $\dfrac{\pi}{3}$ to the right.

Three steps to graph the function $y = -0.2 \sin\left(0.5x - \dfrac{\pi}{6} \right) + 0.1$:

1) We start by sketching $y = -0.2 \sin(0.5x)$ with minimum and maximum values $-0.2$ and $0.2$ over one period $= 4\pi$ (blue graph below).

2) We then sketch $y = -0.2 \sin(0.5x) + 0.1$ by translating the previous graph 0.1 unit up (green graph below).

3) We then shift the previous graph $\dfrac{\pi}{3}$ to the right (red graph below) so that the sketched period starts at $\dfrac{\pi}{3}$ and ends at $\dfrac{\pi}{3} + 4\pi$, which is one period $= 4\pi$.

Graph of y = -0.2 sin(0.5 x - ��/6) + 0.1

Example 5: Working in Degrees

Sketch the graph of $y = 2 \cos(2x - 60^\circ) - 2$ over one period.

Solution:

Graphing Parameters

Amplitude: $|2| = 2$
Period: $\dfrac{360^\circ}{2} = 180^\circ$
Vertical Shift: $-2$ (down 2 units)

Horizontal Shift: Because of the term $-60^\circ$, the graph is shifted horizontally. We first rewrite the given function as:
$$ y = 2 \cos\left[ 2 \left( x - 30^\circ \right) \right] - 2 $$
and we can now write the shift as being equal to $30^\circ$ to the right.

Three steps to graph the function $y = 2 \cos(2x - 60^\circ) - 2$:

1) We start by sketching $y = 2 \cos(2x)$ with minimum and maximum values $-2$ and $+2$ over one period $= 180^\circ$ (blue graph below).

2) We then sketch $y = 2 \cos(2x) - 2$ translating the previous graph 2 units down (green graph below).

3) Finally, we shift the previous graph $30^\circ$ to the right (red graph below) so that the sketched period starts at $30^\circ$ and ends at $30^\circ + 180^\circ = 210^\circ$, which is one period $= 180^\circ$.

Example 6: Fractional Periods

Sketch the graph of $y = -2 \sin\left(\dfrac{x}{3} + \dfrac{\pi}{3}\right) - 1$ over one period.

Solution:

Graphing Parameters

Amplitude: $|-2| = 2$
Period: $\dfrac{2\pi}{\dfrac{1}{3}} = 6\pi$
Vertical Shift: $-1$ (down 1 unit)

Horizontal Shift: Because of the term $\dfrac{\pi}{3}$, the graph is shifted horizontally. We first rewrite the given function as:
$$ y = -2 \sin\left(\dfrac{1}{3}(x + \pi)\right) - 1 $$
and we can now write the shift as being equal to $\pi$ to the left.

Three steps to graph the function $y = -2 \sin\left(\dfrac{x}{3} + \dfrac{\pi}{3}\right) - 1$:

1) We start by sketching $-2 \sin\left(\dfrac{x}{3}\right)$ with minimum and maximum values $-2$ and $+2$ over one period $= 6\pi$ (blue graph below).

2) We then sketch $y = -2 \sin\left(\dfrac{x}{3}\right) - 1$ translating the previous graph 1 unit down (green graph below).

3) We then shift the previous graph $\pi$ to the left (red graph below) so that the sketched period starts at $-\pi$ and ends at $5\pi$ which is one period $= 6\pi$.

$Graph of y = - 2 sin(x/3 + \pi/3) - 1$

Graphing Parameters

Unit Circle Foundations

Sketching Sine and Cosine Functions: Examples with Detailed Solutions

More References and Links