Subject: graphs of other sine & cosine function
Given; 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. If you get this message before Chrism, Merry christmas and the best for the new year. if not best wish anyway. 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 = 0
To graph y = sin(2/3(x-60 period: 540 Phase shift: 60 degree,right
^{o})).If you start with the information period: 540 Phase shift: 60 degree,right
^{o}. The period being 540^{o} means that one period of the sine function takes 540^{o}. Thus this period is completed at 60^{o} + 540^{o} = 600^{o}.
If you have the expression y = sin(2/3(x-60
Since you know the shape of y = sin(t) for t = 0
Cheers, |