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. :!!) :!!)
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].
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).
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.
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)].
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?