Question

Few Sinusodial Function questions.

1. Write an equation for a SINE function with a period of 3pi/2.
2. Write an equation for COSINE a function with a period of 6pi.
3. Write and equation for a function with a period of pi.
4. One cycle of a sine function begins at x= -pi/6
and ends at x= pi/3.
Determine the period of the function.

Answer 1

If you have a function \[y = \sin(x)\]the graph takes a \(2\pi\) range of values of \(x\) to go from its starting point through a complete period to return to the same point. You can see this easily with your unit circle diagram. This means the period has a value of \(2\pi\) radians. Now, if you make it "turn faster" by multiplying the value of \(x\) before giving it to the \(\sin\) function, you get back to the starting point proportionally faster. Multiplying \(x\) by \(2\) will get the argument of \(\sin(2x)\) to go from \(0\) to \(2\pi\) in half the distance along the \(x\)-axis compared to \(\sin(x)\). If we multiply the value of \(x\) by \(2\pi\), we get a rather convenient formulation that allows us to write down sinusoidal functions (sin and cos) with whatever period (or its reciprocal, frequency) we like, or read out the period (or frequency) of a sinusoid we are given.

Answer 2

That formulation is \[y = \sin(\frac{2\pi}{P}x)\] where \(P\) is the period of our sinusoid. In the simple case of the "basic" sine function \[y = \sin(x)\]we can see that we could write that as \[y = \sin(\frac{2\pi}{2\pi}x) = \sin(\cancel{\frac{2\pi}{2\pi}}x) = \sin(x)\]demonstrating that our formulation works for a period of \(P=2\pi\) at the least!

Let's try it out and make a sinusoid with period \(P=\pi\):

\[y = \sin(\frac{2\pi}{P}x) = \sin(\frac{2\pi}{\pi}x) = \sin(2x)\]

Here's a graph of \(y = \sin(x)\) and \(y = \sin(2x)\) for comparison:

Answer 3

You can see if you look at the graph that \(y = \sin(x)\) goes through one complete cycle in the space of \(2\pi\) along the \(x\)-axis, and \(y = \sin(2x)\) goes through two complete cycles in the same space, or one complete cycle in \(\pi\), just as we set out to do.

Answer 4

The cosine function works the same way; it's really just a sine function shifted along the \(x\)-axis a bit.

Finding the period of a function is easy. Say you have \[y = \sin(\frac{23}{7}x)\]which I think you'll agree is a bit ugly. What you do is take the argument of the \(\sin\) function and set it equal to \(\frac{2\pi}{B}x\) and solve for the value of \(B\).

\[\frac{23}{7}x = \frac{2\pi}{B}x\]we can cancel out the \(x\)'s as they appear on both sides, leaving us with
\[\frac{23}{7}=\frac{2\pi}{B}\]now we cross-multiply:
\[23*B = 7*2\pi\]divide both sides by \(23\) and we have\[B = \frac{14\pi}{23}\]
The period of that function is \(\frac{14\pi}{23}\approx 1.91227\)

And here's the graph:

Answer 5

For the final part of your problem, you need to figure out the length of the segment of the \(x\)-axis between \(x=-\frac{\pi}{6}\) and \(x = \frac{\pi}{3}\). Once you know that, you know the period.