Graphs of Sine and Cosine FunctionsI.
Sine Curve: The graph of the sine function is a sine curve.
Recall: the period of the sine function is 2π and the sine function is an odd function.
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On the unit circle, the sine function was the y-value. If we cut the unit circle at the 0, 2π place, and used the angle measure as the x-value and the sine function (or y-values from the unit circle) stays as the new y-values, you will get the sine curve.
Some of the points are (0, 0), (π / 2, 1), (π, 0), (3 π / 2, -1), (2 π, 0).
The sine curve is symmetric to origin.
Domain: all the real numbersRange: [-1, 1]
II. Cosine curve:The graph of the cosine function is the cosine curve.
Recall: The period of the cosine function is 2 π and the cosine function is an even function.
Some of the points are (0, 1), (π / 2, 0), (π, -1), (3 π / 2, 0), (2 π, 1).
The cosine curve is symmetric with respect to the y-axis.
Domain: all real numbersRange: [-1, 1]
To sketch the graph: Note 5 key points; maximums, minimums, and intercepts.
Divide the period into four equal parts to get the key points.
Amplitude:The amplitude of y = a sin x and y = a cos x represents half the distance between the maximum and minimum values of the functions and is given by:
Amplitude a or (1/2 (maximum value - minimum value)
Example 1: y = 32 sin x
a = 32
Period = (2 π)/ 1 = 2 π
Dividing 2 π into 4 equal parts, you get the points(0,0), (π / 2, 1), (π, 0), (3 π / 2, -1), (2 π, 0)
Recall: y = - f(x) is a reflection in the x-axis of the graph of y = f(x)
Example 2: y = -32 sin x
It is the same as example 1 only reflected in the x-axis
so the points are(0,0), (π / 2, -1), (π, 0), (3 π / 2, 1), (2 π, 0)
Period of Sine and Cosine Functions
Let "b" be a positive real number. The period of y= a sin bx and y = a cos bx is given by:
Period = (2 π) / b
If b is between 0 and 1 then horizontal stretch
If b is less than 1 then horizontal shrink
If b is negative then use:sin(-x) = - sin x and cos (-x) = cos (x)
Example 1: y = 2 cos (3x)
a = 2
period =(2π / 3)
(2 π / 3) / 4 = π / 6
so the points are:(0, 2), (π / 6, 0), (π / 3, -2), (π / 2, 0), (2π / 3, 2)
Example 2: y = (3/2) sin (π / 2 x)
a = 3/2
period = (2 π)/ (π / 2) = 4
So the points are:(0, 0), (1, 3/2), (2, 0), (3, -3/2), (4, 0)
Example 3: y = -2 cos (12π x)
a = 2
period = (2π)/ (12π) = 1/6
so dividing 1/6 by 4 you get 1/24
so the points are:(0, -2), (1/24, 0), (1/12, 2), (1/8, 0), (1/6, -2)
Example 4: y = 3 sin (-3x)
Recall y = sin (-x) = -sin x
so we would havey = -3 sin(3x)
a = 3
Period = (2π)/ 3
so dividing that into 4 equal parts
you will get the points:(0, 0), (π / 6, -3), (π / 3, 0), (π / 2, 3), (2π / 3, 0)
Homework: #31 pg 330; # 1- 13 odd, 17, 19, 27, 39, 41, 43, 49, 71, 73, 75
pg. 320; #9, 11, 25, 32
