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.
Can we see the statement plz
Answer:
The answer is b
pls brainleist
Step-by-step explanation:
The answer is C: first add 2 both sides then divide both sides by -5.
