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 range is [-1, 1].
There are infinitely many zeros, each of which is located at the midpoint
between the nearest maximum and minimum.

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)		ymax = 3	ymin = -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)		ymax = 2	ymin = -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 b doesn't effect the curve the same way it does for a sine curve.

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 + o /8) + 3.
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 + o /8) + 3 becomes f(x) = 4 sin ¼ ( x + o/2) + 3

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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