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Difference between Cos and sin

by tiffney
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tiffney
#1
Apr26-05, 11:46 AM
P: 3
Can anyone tell me the difference btween cos and sin, and tell me when creating waveform why the sin function creates a sawtooth appearance and the cos function does. Any help would be appreiciated.
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Data
#2
Apr26-05, 12:49 PM
P: 998
Cosine is a phase shift of sine (and visa-versa). In other words,

[tex]\sin \theta = \cos \left(\theta - \frac{\pi}{2}\right)[/tex]

just a translation to the right by [itex]\pi / 2[/itex].
Icebreaker
#3
Apr26-05, 12:50 PM
P: n/a
Given a right triangle and a non-right angle, the sine of that angle is equal to the length of the side facing the angle over the length of the hypotenuse. The cosine of the said angle is equal to the length of the side adjacent to the angle over the length of the hypotenuse.

A cosine wave is simply a sine wave that is shifted to the right by pi/2.

It would be much easier to explain if I can draw pictures. Your best chance is to google it or to check out MathWorld (http://mathworld.wolfram.com).

mathwonk
#4
Apr26-05, 01:13 PM
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Difference between Cos and sin

for a point on the unit circle, cos is the x coordinate and sin is the y coordinate. I.e. one is the shadow of the radius on the x axis and the other the shadow of the radius on the y axis. Due to the symmetry of the circle as you go around the length of the shadow of the radius on the x axis or on the y axis look essentially the same, just out of phase.
BobG
#5
Apr26-05, 03:06 PM
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Quote Quote by mathwonk
for a point on the unit circle, cos is the x coordinate and sin is the y coordinate. I.e. one is the shadow of the radius on the x axis and the other the shadow of the radius on the y axis. Due to the symmetry of the circle as you go around the length of the shadow of the radius on the x axis or on the y axis look essentially the same, just out of phase.
And the y-axis is perpendicular (90 degrees; pi/2 radians) to the x-axis. Which is why:

Quote Quote by Data
Cosine is a phase shift of sine (and visa-versa). In other words,

[tex]\sin \theta = \cos \left(\theta - \frac{\pi}{2}\right)[/tex]

just a translation to the right by .
An Average Joe
#6
Apr29-05, 10:37 PM
P: 18
The "difference" between cos(x) and sin(x) is |cos(x) - sin(x)|.






That was a joke.






Really.
jtbell
#7
Apr30-05, 12:02 AM
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Quote Quote by tiffney
when creating waveform why the sin function creates a sawtooth appearance and the cos function does.
Eh, what?
JamesU
#8
Apr30-05, 12:06 AM
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I suggest google, it's not the easiest thing to explain
Data
#9
Apr30-05, 12:08 AM
P: 998
several people have explained it here already
whozum
#10
Apr30-05, 03:02 AM
P: 2,218
Not his sawtooth conjecture, which beats me.
dextercioby
#11
Apr30-05, 10:51 AM
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The only thing that connects sawteeth & sin/cos is Fourier series/transform...

Daniel.
robphy
#12
Apr30-05, 12:29 PM
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An important property of sin(x) and cos(x) is that
cos(x) is an "even function of x" [that is, cos(-x)=cos(x)] and
sin(x) is an "odd function of x" [that is, sin(-x)=-sin(x)].
The Reverend BigBoa
#13
Apr30-05, 09:13 PM
P: 20
I'm thinkin' something's missing here.

No cos or sin function, at least involving perhaps your basic constant coefficients, generates a sawtooth wave, does it? It should generate a sine wave and another wave 90 degrees out of phase. Perhaps his problem involves some Fourier or other transfoms as Dexter already suggested?


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