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Period of Sine and Cosine

thunc14

New member
Joined
Nov 15, 2017
Messages
22
I understand that the period for sine and cosine functions are theta (in radians) + 2pi.

But if you have a point on the unit circle in the first quadrant, and you drew a line parallel to the x-axis until you reached the unit circle in quadrant 2, wouldn't that be the same y-value, thus the same value for sin theta?

If you repeated the process from the point in quadrant 1 and drew a line parallel to the y-axis until you reached the unit circle in quadrant 4, wouldn't that be the same x-value, thus the same value for cos theta?

I don't understand how this works, if the period is 2pi, then why are there other points on the unit circle that would give the same value for x and y, therefore sine and cosine respectively?

I attached a picture, hopefully that makes my confusion clearer.
Screenshot 2018-01-23 at 14.32.16.jpg
 

mmm4444bot

Super Moderator
Staff member
Joined
Oct 6, 2005
Messages
10,143
I understand that the period for sine and cosine functions are theta (in radians) + 2pi.

But if you have a point on the unit circle in the first quadrant, and you drew a line parallel to the x-axis until you reached the unit circle in quadrant 2, wouldn't that be the same y-value, thus the same value for sin theta?
Yes. It would be the same value.


If you repeated the process from the point in quadrant 1 and drew a line parallel to the y-axis until you reached the unit circle in quadrant 4, wouldn't that be the same x-value, thus the same value for cos theta?
Yes. It would be the same value.


I don't understand how this works, if the period is 2pi, then why are there other points on the unit circle that would give the same value for x and y, therefore sine and cosine respectively?
If you take a graph of sin(x) or cos(x) plotted, say, from x = -4·Pi to x = 4·Pi, and then drew a horizontal line through it, say y=1/2, you'll see that this line intersects the graph at two locations, within any interval of width 2·Pi.

Another way of thinking about it is with respect to the symmetry of a circle. As the terminal ray of the angle (in standard position with a unit circle) rotates about its vertex, the point of its intersection with the circle is traveling around the circumference in a counterclockwise direction. For y=sin(x) as angle x increases from x=0 to x=Pi/2 radians, this point is rising in Quadrant I (the y-coordinate is increasing from 0 toward 1).

Once angle x has increased past Pi/2, the point moves past the point (0,1) and starts descending in Quadrant II toward the point (-1,0). Due to the symmetry of the circle, the point's y-coordinate will pass through every value on its way back to zero that it just finished passing on its way up in the first quadrant. That is, sin(x) takes on every value between 0 and 1 twice, as the terminal ray rotates through Quadrant I and Quadrant II.


I think the best thing for you to do at this point, is to google keywords sine cosine unit circle animation, and check out a lot of the animated .GIFs, videos, interactive geometry or computer algebra sites, and watch all the different graphical representations of the periodicity of the sine and cosine functions.

After that, if you still have questions about why sin(x) and cos(x) values repeat within one period, let us know. :cool:

Here's one short video showing sin(x), but do check out a lot more, by googling.


[video=youtube;Ohp6Okk_tww]https://www.youtube.com/watch?v=Ohp6Okk_tww[/video]
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
2,760
I understand that the period for sine and cosine functions are theta (in radians) + 2pi.
I think you mean that the period is 2 pi, which means that for any theta, sin(theta + 2 pi) = sin(theta), and similarly for cosine.

But if you have a point on the unit circle in the first quadrant, and you drew a line parallel to the x-axis until you reached the unit circle in quadrant 2, wouldn't that be the same y-value, thus the same value for sin theta?

If you repeated the process from the point in quadrant 1 and drew a line parallel to the y-axis until you reached the unit circle in quadrant 4, wouldn't that be the same x-value, thus the same value for cos theta?

I don't understand how this works, if the period is 2pi, then why are there other points on the unit circle that would give the same value for x and y, therefore sine and cosine respectively?
The fact that sin(theta + 2 pi) = sin(theta) for any theta does not imply that theta + 2 pi is the only other angle that produces the same sine. This is an "if ... then" situation: it must be read in only one direction. Similarly, we can say that if an animal is a cat, then it has fur; but that does not mean that a cat is the only animal that has fur, or that if an animal has fur, then it is a cat. So the fact that the sine is periodic doesn't require that there can be no other values within a period that have the same sine.
 

thunc14

New member
Joined
Nov 15, 2017
Messages
22
Perhaps I confused the definition of the period of a trig function to mean what radian angle would get you the same trig value (in this case the same y values for sine or same x values for cosine), rather than when the function begins to repeat itself.
 

j-astron

Junior Member
Joined
Jan 10, 2018
Messages
181
Perhaps I confused the definition of the period of a trig function to mean what radian angle would get you the same trig value (in this case the same y values for sine or same x values for cosine), rather than when the function begins to repeat itself.
Yup.

Say you have some function \(\displaystyle f(\theta)\). The formal definition of the period of a periodic function is that it's the value T such that \(\displaystyle f(\theta + T) = f(\theta)\) for all \(\displaystyle \theta\). The words "for all" are key there. It means that no matter which theta-value you start at along the horizontal axis, if you move +T away from it, you'll return to the same y-value on the curve as you started with.

What you basically did in your original post was the following:

1) you looked at your function \(\displaystyle f(\theta) = \cos(\theta)\)

2) You drew a point at \(\displaystyle \theta = \pi/4 \) and noticed that \(\displaystyle \cos(\pi/4) = \sqrt{2}/2\)

3) You drew another point at \(\displaystyle \theta = -\pi/4 \) and noticed that \(\displaystyle \cos(-\pi/4) = \sqrt{2}/2\) as well

4) Since the cosine value repeated between these two angles, you wondered why the period isn't just \(\displaystyle \pi/4 - (-\pi/4) = \pi/2 \)

Well, let's check that. Is the period of cosine really pi/2? Suppose we start at a different theta value and move away from it by pi/2. For example, suppose we start at theta = 0

\(\displaystyle \cos(0) = 1\)

\(\displaystyle \cos(0+\pi/2) = \cos(\pi/2) = 0\)

0 is not equal to 1, so this time the cosine value didn't repeat when we moved by an interval of pi/2. So clearly pi/2 is not the period of the function.

Basically the period is the shortest distance over which the function is guaranteed to always repeat, regardless of where you start on it. For some starting values, the function may repeat over shorter intervals, but it won't be true for all starting points.
 
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