Graphs of Sine and Cosine Curves |

**Graphing Sine Curves**

**The Basic Sine Curve:*** f(x) = sin x*.

Since radian measure represents length as well as rotation (angles),

we express our **x-values** in radians and our y-values in *sin x* values.

Using this approach, we can graph trig functions.

From the *x *and* y values *of the* unit circle* we know:

sin 0 = 0 |
sin o /2 = 1 |
sin o = 0 |
sin 3o /2 = -1 |

Here, we see two cycles of **f(x) = sin x**.

This pattern will repeat to the right & left indefinitely.

As you can see, the **domain** is ** R**, the

There are

The properties of the cyclical trig graphs still include all the elements of the former list, (ie: domain, range, max, min, increasing, decreasing and signs) but it also includes:

**amplitude, period, frequency, phase shift and vertical translation.**

*(see lesson file tr4.2 properties of sinusoidal curves for information)*

**Example1:** Let's graph *f(x) = 2 sin 4(x +**o**/2) + 1*

amplitude: 2 |
period: | 2 o/ 4 | = o/2 |
phase shift: -o/2 |
vertical transl: 1 |

starting point = (-o/2, 1) |
y_{max} = 3 |
y_{min} = -1 |

As we see from the graph, the line **y = 1** forms a horizontal axis for the curve.

We find the starting and the ending point, of a cycle, divide the interval

into 4 equal parts, then sketch the curve.

** Note:** To divide the interval into 4 equal parts, find the starting and the ending point, find the midpoint of those, then find the remaining 2 mid points.

**Example 2:** Let's graph one more sine curve.

Graph ** ***f(x) = -4 sin 3(x -**o**/3) - 2*

amplitude: 4 |
period: | 2 o/ 3 | = 2o/3 |
phase shift = o /3 |
vertical transl: -2 |

starting point = (o/3, -2) |
y_{max} = 2 |
y_{min} = -6 |

since **a** < 0, the curve will decrease to its min value before rising to a max.

starting point = **(****o****/3,** **-2), ** period = **2**** ****o****/3,** so ending point = **o****/3 + 2**** ****o****/3 = ****o**** .**

__Note__:**2 ****o****/3** = **4 ****o****/6** so each quarter of the period will =**o****/6**, so we just add** ****o****/6** to each x-value

The midpoint between the start and end is

(o/6 + o/3 = o/2 ) and (o/6 + 2 o/3 = 5 o/6).

The vertical translation is -2, so the curve's horizontal axis is the line y = -2.

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**Graphing Cosine Curves :**

The curve of a cosine function is **isomorphic** to a sine function's curve that has been pulled back, or phase shifted** ****o****/2** units to the left. In other words, the cosine curve starts at its maximum or minimum value. Let's look at **y = cos x** and then we'll discuss the properties of the function.

As you can see, the curve begins at the maximum if **a** > 0 .

If **a** < 0 the curve starts at the minimum.

The period, amplitude, phase shift and vertical translation are identical to those of the sine curve.

*(see lesson file tr4.2 properties of sinusoidal curves for information)*

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Let's graph **y = -3 cos 2(x + ****o****/3) - 1**.

amp: 3 |
period: o |
phase shift: -o/3 |
vert. trans: -1 |

max y value = 2 |
min y value = -4 |
start point: (-o/3, -4) |
end point: (2o/3, -4) |

*** Note:** the parameter

Since **cos x** is positive ( > 0) in the 1st and 4th quads, when **b** is negative, it doesn't change the shape or starting point for the cosine curve.

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**Practice**

1/ Graph one cycle of the function *f(x) = 4 sin ( ¼ x + *

List the amplitude, period, phase shift, maximum, minimum and vertical translation.

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2/ Graph one cycle of the function *g(x) = - 3 cos (2x + **o**) - 5*.

List the amplitude, period, phase shift, maximum, minimum and vertical translation.

(*for more practice see lesson file tr536qz: trig quiz in the Trig MathRoom)*

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**Solutions**

1/ *f(x) = 4 sin ( ¼ x + *

amp: 4, up first |
period: 8o |
phase shift: -o/2 |
vert. trans: 3 |

max: 7 |
min: -1 |
start point: (-o/2, 3) |
end point: (15o/2, 3) |

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2/

amp: 3 |
period: o |
phase shift: -o/2 |
vert. trans: -5 |

max y value = -2 |
min y value = -8 |
start point: (-o/2, -8) |
end point: (-o/2, -8) |

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