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Transformations of Graphs of Trigonometric Functions

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Describe the transformations required to obtain the graph of the given function from a basic trigonometric graph.
43. y=0.5 sin 3x
How do you come up with the following answer?
starting from y=sinx, horizontally shrink by 1/3 and vertically shrink by 0.5
45. y= -2/3 cos x/3

How do you come up with the following answer?
starting from y=cos x, horizontally stretch by 3, vertically shrink by 2/3, reflect (?) across x-axis.

Describe the transformations required to obtain the graph of y2 from the graph of y1.
49. y1=cos2x and y2=5/3cos 2x
how do you come up with the following ans.
starting from y1, vertically stretch by 5/3

Select the pair of functions that have identical graphs

53. a) y = cos x b) y= sin(x + pie/2)
c) y = cos (x+pi/2)

how do you come up with a & b as the ans.

Construct a sinusoid with the given amplitude and period that goes through the given point.
57. amplitude 3, period pie, point (0,0)
anwer
y=3 sin 2x (How do you get 2x?)

61. state the amplitude and period of the sinusoid, and (relative to the basic function) the phase shift and vertical translation
y= -2 sin (x - pi/4) + 1
Answer: How do you know phase shift is pie/4 and vertical translation is 1 unit up.

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Transformations of graphs of trigonometric functions are investigated.

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43. y = 0.5 sin 3x
The standard curve is: y = A sin ,
where A = amplitude
T = time period.
Now look at the equation of the standard sine curve y = sin x, which can be rewritten as y = 1*sin . When you compare this equation to y = A sin , we get A = 1, T = 2. So the amplitude of this equation is 1 and the time period is
Now for the equation y = 0.5 sin 3x, when you compare this equation to y = A sin , we get A = 0.5 and T = . Therefore, when compared to y = sin x, the amplitude of this equation should be halved and the time period should be reduced to one-third. This is how we get the answer: starting from y = sinx, ...

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