If you were to graph the sine and cosine functions on the same set of axes, you'd see that they are 90 degrees, or pi/2 radians, out of sync with one another. cos 0 is 1, whereas sin 0 is 0; sin x does not reach the value 1 until your angle reaches 90 degrees, or pi/2 radians.
Please do some experimentation here. You want to express sin (300t) as a cosine, that is, as cos (300t + [some angle]), where [some angle] is called a "phase shift."
Start with the basic y=sin x. Its graph is usually begun at (0,0). Try simplifying and graphing cos (x-pi/2). Does this produce the same y=sin x, with the same graph? Do you remember that cos (x-pi/2) = cos x cos pi/2 + sin x sin pi/2? It happens that cos pi/2 = 0 and that sin pi/2 = 1. Thus, cos (x-pi/2) = sin x (1) = sin x. So, we have succeeded in obtaining sin x from cos (x-pi/2).
Now, what about obtaining sin 300t from the cosine function?
First: recognize that the standard form of the cosine function with a phase shift is y = a cos (bx + c). What is the period?
Answer: The period is always 2pi/b. So, in the case, the period is 2pi/300, or pi/150.
What is the phase shift? Answer: the period is always -c/b. So, in this case, the period is -c/b, or -pi/2 over 300. This simplifies to -pi/150.
Try this: Simplify cos (300t -pi/150) If the end result is sin 300t, you'll know you have this right. If the end result is not sin 300t, experiment with that phase shift.
