Subject: graphs of other sine & cosine function

Given;
amplitute:1
period: 540
Phase shift: 60 degree,right

I am ask to right the equation: sin 2/3 (value -60degree)

When I am asked to graph the equation, the period is mixing me up.

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pierre

Hi Pierre,

To do this you need to keep in mind the graph of y = sin(t) for one period of the sine function, t = 0o to t = 360o.

To graph y = sin(2/3(x-60o)) you can either use the given information:

amplitute:1
period: 540
Phase shift: 60 degree,right

 or the expression y = sin(2/3(x-60o)).

If you start with the information

amplitute:1
period: 540
Phase shift: 60 degree,right

 then the phase shift information tells you to start the sine graph at +60o. The period being 540o means that one period of the sine function takes 540o. Thus this period is completed at 60o + 540o = 600o.

If you have the expression y = sin(2/3(x-60o)) and not the information about period and phase shift then you can plot the graph as follows.

Since you know the shape of y = sin(t) for t = 0o to t = 360o solve 2/3(x-60o) = 0 to get x = 60o and 2/3(x-60o) = 360o to get x = 600o. Thus the phase shift is 60o and the graph goes through one cycle from 60o to 600o. Hence its period is 600o - 60o = 540o.

Cheers,
Harley

 

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