Finding the period for a sine or cosine function

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bro123

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Hi everyone, I am having trouble finding the period for a sine or cosine function like the one shown below, how would you go about finding the period for this graph? Appreciate the help!
Screen Shot 2020-12-10 at 9.04.19 pm.png
 
Subhotosh Khan said:
First calculate 'c'
Click to expand...
so that would be -4 right and so would that mean that i then find 'a' which would be 6 and the do i possibly sub in a point on the graph to then find 'b' which would be the period? or am i missing something?
 
You could find b by substituting in a point, but consider this way:
The period can be found from the graph The period is how long it takes for one cycle of the graph. For example, look at the point (0, -4). How long (in the x-direction) does it take before the graph repeats itself? Trace your finger along the curve from (0, -4) to (2.5, -4). The graph then repeats itself, so the period is 2.5.
Also the period can be found from the equation. The period is the coefficient of x divided by 2pi. In this case the period is b*pi/(2pi). So equate this to 2.5 to find b.
Let us know what you get.
 
Harry_the_cat said:
The period is the coefficient of x divided by 2pi. In this case the period is b*pi/(2pi).
Click to expand...

I think that the quote above might be the way to obtain frequency rather than period. Period and frequency are just reciprocals of each other (period=1/freq) so the OP can still use the good advice above, with a slight adaption, to obtain the correct answer. Period of the equation is (2pi)/(b*pi). Equate this to 2.5 to find b.
 
Here is how you find the period. Pick any point, I suggest that you pick a point that is either the starting point of a sine or cosine graph. Now find the end of that period. Compute the absolute value of the difference in the two x-values and this is your period.
 
Jomo said:
Here is how you find the period. Pick any point, I suggest that you pick a point that is either the starting point of a sine or cosine graph. Now find the end of that period. Compute the absolute value of the difference in the two x-values and this is your period.
Click to expand...
Thou hast not fully digested the content of post#4, wherein these gems of knowledge are already contained :D
 
@bro123 note that you can check your final solution very quickly by plugging some x values into your f(x) formula. I'd suggest x=0, x=2.5, x=2.5/4 (you'd expect the value 2 here). However please post back with your work!
 
Cubist said:
I think that the quote above might be the way to obtain frequency rather than period. Period and frequency are just reciprocals of each other (period=1/freq) so the OP can still use the good advice above, with a slight adaption, to obtain the correct answer. Period of the equation is (2pi)/(b*pi). Equate this to 2.5 to find b.
Click to expand...
Yes you are correct - my mistake. Thanks Cubist!
 
Cubist said:
@bro123 note that you can check your final solution very quickly by plugging some x values into your f(x) formula. I'd suggest x=0, x=2.5, x=2.5/4 (you'd expect the value 2 here). However please post back with your work!
Click to expand...
Harry_the_cat said:
You could find b by substituting in a point, but consider this way:
The period can be found from the graph The period is how long it takes for one cycle of the graph. For example, look at the point (0, -4). How long (in the x-direction) does it take before the graph repeats itself? Trace your finger along the curve from (0, -4) to (2.5, -4). The graph then repeats itself, so the period is 2.5.
Also the period can be found from the equation. The period is the coefficient of x divided by 2pi. In this case the period is b*pi/(2pi). So equate this to 2.5 to find b.
Let us know what you get.
Click to expand...
ok so I worked it out to be f(x)=6*sin(0.8*pi*x)-4 and I plugged in points as suggested and I believe it is correct
thanks everyone for the help!!!
 
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