Current Events > Anyone good at Sine and Arcsine (Topic Part 3)

(+) Yeah! (+)

Spending hours trying to understand the course material and it's not clicking at all.

Circle has a radius of 105, the period is (0.9, midline at 135. I have "f(x) = 105sin(0.9x)+135".

For the inverse, I have "y = sin^-1 ((f[x]/105))"

Except when I do that, I get these weird disconnected semi-circles that never intersect with the original graph at all... It just doesn't look right but I can't figure out what the issue is.

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Could you post a picture of the problem you're trying to solve?

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emmo posted...

Could you post a picture of the problem you're trying to solve?

well, I gave all the information I have.

The green semicircles are what I get when I enter that inverse function.

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Hmm. Are you trying to obtain a graph of the corresponding arcsine function and superimpose it?

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emmo posted...

Hmm. Are you trying to obtain a graph of the corresponding arcsine function and superimpose it?

Yes. As I understand it, the output (y) for the arcsine function is supposed to equal the angle measurement of the corresponding point on the original sine function, and it covers Q1 and Q4 only.

But this "function" I have only covers Q1... it doesn't seem like much of a function since it just loops back up on itself and never goes below 0 (to give me any Q4 data).

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So the first part is a straightforward A(x) = 3 sin (x). When I type in the inverse function of that, I have y = sin^-1 (A[x]/3) and it gives me what you would expect to see:



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I had to include a -135 in my arcsine function... no idea why (to negate the +135 of the original? But it's just the midline, doesn't seem like that should play a role).

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still not entirely sure it's correct since everything I've seen about arcsine functions except for one video is that it's meant to be a single segment that travels along the y-axis from 2pi to -2pi and that's it.

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That isn't an arcsine graph. What you have in this second example is a linear graph made periodic by taking the inverse sine of a sine. If that doesn't make sense, consider the following line:

A(x) = 3 sin[x] --> A(x)/3 = sin[x] --> y(x) = arcsin[A(x)/3] = arcsin[sin[x]] = x --> y(x)=x

The slope switches direction because the computation is switching signs at pi/2, 3pi/2, etc.

An arcsine graph should be curved, and to be a function, you must restrict the domain to an appropriate interval. The domain of the arcsine function will simply be the range of the corresponding sine function, and vice-versa. Since you're looking for Q4 and Q1 data, the most straightforward domain for the sine function would then be -pi/2 < x < pi/2. In the case above, this corresponds to a range of -3 < y < 3. Switching the variables will give the domain of the arcsine function, then, as -3 < x < 3. Finally, you need to be careful about how you set up the arcsine function, as the total argument (or input) can only accept values between -1 and 1. This is because the output of a sine operator is only ever between -1 and 1, so this is the only acceptable input to an arcsine. Thus, in this example, the arcsine function should look like this:

y(x) = sin^-1 [x/3] , -3 < x < 3

Plug this in and you'll get a graph that looks much different.

The same logic applies to the problem at hand. The first equation looks fine except for the argument in the sine function. I say this because you identified the period as 0.9, but used this value where the angular frequency should be. This is an easy fix, as the period T is related to the angular frequency w by the relation:

w = 2pi/T

whereas the general form of an oscillating function (without phase) is:

f(t) = A sin[wt] + C , C = constant

Of course, you can swap out the variables for the purpose of plotting this without changing anything. The addition of the constant 135 at the tail end translates the graph up the y-axis to the chosen midline, so no problem there. Just note that you cannot include this constant in the argument of the arcsine function, for the reason already discussed.

So, changing our variables and rearranging, we have:

y = A sin[wx] + C --> (y - C)/A = sin[wx] --> wx = sin^-1 [(y-C)/A] --> x = (1/w) sin^-1 [(y-C)/A]

Once you plug in your values, swap the variables for the purpose of plotting, and define an appropriate domain, you should be good to go. Note that this will be centered about the origin, so you'll need to add a constant if you wish to translate it such that both graphs share a midline.

Hope that's helpful.

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By inverse here, do you just mean if you have y = f(x), you want x = f(y)?

Hexenherz posted...

I had to include a -135 in my arcsine function... no idea why (to negate the +135 of the original? But it's just the midline, doesn't seem like that should play a role).

This suggests the above is the case. You are algebraically solving for y in terms of x. So if you do the algebra for that, you should get that answer.

Hexenherz posted...

still not entirely sure it's correct since everything I've seen about arcsine functions except for one video is that it's meant to be a single segment that travels along the y-axis from 2pi to -2pi and that's it.

arcsin or sin^-1 will give you a sine curve that has been flipped and rotated from the x axis.

For y = sin(x), 1 = sin(pi/2), -1 = sin(-pi/2).

for x = sin(y), pi/2 = arcsin(1), -pi/2 = arcsin(-1)

As you can see, x and y are the same across these functions, but their position has been moved around.

Now let's think about y = arcsin(x). What that means effectively is a sine curve on the Y-axis. In theory, that would be an infinitely repeating wave going up and down. However, a function cannot have two y values for a single x value (but the same value can occur at multiple x values). Because of this, arcsin or arccos functions only exist in a range in which no values of y repeat.

So there is only one full cycle between -1 and 1 on the x axis that is represented for arcsin and arcos. For arcsin that is between -pi/2 and pi/2, for arccos it is 0 to pi in typical graphing software without transformations. Note that you yourself could choose a different y domain - you should simply choose the cycle that best works for you.

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I never cared for the arcsin and arccos functions. At least the arctan function has some use for integration and converting to polar coordinates.

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Thank you all for the information, gonna save it to review later. Read through it but I'll need to take notes and play around with it to really understand it. Already spent way too many hours on this one assignment -_-

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