# Graphs of Sine and Cosine

The sine curve has many applications in Physics. For example, sound, light and electromagrnetic waves all travel in a sine wave.

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The graphs of $y = sin(x)$ {sometimes called the sine wave} and $y = \cos(x)$ are derived from the definition of sin and cos using the unit circle:

- If P is a point on the unit circle measured an angle of $\theta$ from the positive x-axis,
- then $\sin(\theta)$ is defined as the y-coordinate of P.
- and $\cos(\theta)$ is defined as the x-coordinate of P.

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For all trig graphs,

- the
**median line**(or mean line) is a horizontal line through the centre of the graph - the
**amplitude**is the maximum height above the median line - the
**period**is the distance in the x-direction to complete one cycle of the graph

… … Source: Wikipedia (Sine)

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## Graph of y = sin(x)

The above diagram only shows one cycle of sine. The sine function actually continues to infinity.

For the graph, $y = \sin(x)$

- The median line is along the x-axis ($y = 0$)
- The amplitude is a = 1
- The period is $2\pi$
- Domain: $x \in R$
- Range: $y \in [ -1, \; 1 ]$

## Graph of y = cos(x)

The graph, $y = \cos(x)$ is the same shape as $y = \sin(x)$ but translated $\frac{\pi}{2}$ to the left.

For the graph, $y = \cos(x)$

- The median line is along the x-axis ($y = 0$)
- The amplitude is $a=1$
- The period is $2\pi$
- Domain: $x \in R$
- Range: $y \in [-1, \; 1 ]$

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## Transformations

We can use the standard transformations, such as dilations, translations and reflections on the trig graphs.

### Dilations in the Y-Direction (from x-axis): $y = a\sin(x)$

$y = a\sin(x)$ causes a dilation by a factor of a on the x-axis (in the y-direction),

- in the same way that $y = af(x)$ causes a dilation on the x-axis.

Because the amplitude of $\sin(x)$ is **1**, a dilation by a factor of a changes the amplitude to **a**.

Recall that if **a** is negative, we get a reflection across the x-axis (in the y-direction).

**Note:** Since amplitude is always positive, the amplitude is the magnitude of a $|a|$.

#### Example 1

Sketch the graph, $y = 2\sin(x)$

**Solution**

- The median line is along the x-axis ($y = 0$)
- Amplitude $a = 2$
- Period: $2\pi$
- Domain: $x \in R$
- Range: $y \in [-2, \; 2]$

The x-intercepts can be found by solving $2\sin(x) = 0$

- x-intercepts: $x = k\pi, \quad k \in Z$

{we use k instead of n because n has a different meaning here}

## Dilations in the X-Direction (from y-axis): $y = \cos(nx)$

$y = \cos(nx)$ causes a dilation by a factor of $\frac{1}{n}$ in the x-direction, in the same way that $y = f(nx)$ causes a dilation.

Because the period starts at $2\pi$, a dilation by $\frac{1}{n}$ changes the period to $\frac{2\pi}{n}$

Recall that if **n** is negative, we get a reflection across the y-axis (in the x-direction).

Note: Since period > 0, the period is $\dfrac{2\pi}{|n|}$

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#### Example 2

Sketch the graph, $y = \cos(2x)$

**Solution**

- The median line is along the x-axis ($y = 0$)
- Amplitude $a = 1$
- Period: $\frac{2\pi}{2} = \pi$
- Domain: $x \in R$
- Range: $y \in [-1, \; 1]$

The x-intercepts can be found by solving $\cos(2x) = 0$

- x-intercepts: $x=k\pi \pm \dfrac{\pi}{4}, \; k \in Z$

or, more simply:

- x-intercepts: $x= \left( 2k+1 \right) \dfrac{\pi}{4}, \; k \in Z$

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#### Example 3

Sketch the graph $y=\sin \left( x-\dfrac{\pi}{2} \right)+2$

**Solution**

- Median line: $y = 2$
- Amplitude:
**1** - Period: $2\pi$
- Domain: $x \in R$
- Range: $y \in [1, \; 3]$

The points on the median line can be found by solving $y = 2$.

- median points: $x = \left( 2k+1 \right) \dfrac{\pi}{2}, \; k \in Z$

- x-intercepts: none … (found by solving $y = 0$ )

## Summary

$y = a\sin \big(n(x - b) \big) + c$ {and same for cos}

- dilated by a factor of a in the y-direction
- dilated by a factor of $\frac{1}{n}$ in the x-direction
- if a < 0, reflected in the y-direction
- if n < 0, reflected in the x-direction
- translated b units to the right
- translated c units up

SO:

- Median line: $y = c$
- Amplitude = $|a|$
- Period = $\frac{2\pi}{n}$
- Domain: $x \in R$
- Range: $y \in [c + a, \; c - a]$
- Median points: solve $y = c$
- x-intercepts: solve $y = 0$

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