# Finding inverse of a Sin function (problem from Mooculus)

• Ryaners
In summary: Hallo again, Ray - I think the error in my approach is obvious now. I don't think h(t) is actually invertible on the restricted domain [3.5, 10.5]. I'm not sure how I missed that - I had it in my head that the Ferris wheel would be at the same height twice on each ride, but that is of course impossible with a sinusoidal function. It just goes to show how dangerous it is to think you can solve a problem you haven't fully understood. Thanks for your help!In summary, the student attempted to find the inverse of a function representing the height of a
Ryaners
I'm working through the problems in the Mooculus textbook as revision for Calculus I & there seems to be something wrong with how I'm manipulating the function to find its inverse in the following example.

## Homework Statement

The height in meters of a person off the ground as they ride a Ferris Wheel can be modeled by:
h(t) = 18 · sin[(π · t )/ 7] + 20
where t is time elapsed in seconds. If h is restricted to the domain [3.5, 10.5], find and interpret the meaning of h-1(20).

## The Attempt at a Solution

Here are my workings:

Let y = h(t)
y = 18⋅Sin[(π⋅t)/7] + 20
(y - 20) / 18 = Sin[(π⋅t) / 7]
arcsin [(y - 20) / 18] = (π⋅t) / 7
π⋅t = 7⋅{arcsin [(y - 20) / 18]}
t = (7/π)⋅{arcsin [(y - 20) / 18]}

⇒ h-1(t) = (7/π)⋅{arcsin [(t - 20) / 18]}
⇒ h-1(20) = (7/π)⋅arcsin(0) = 0

However this is the solution given:
h-1(20) = 7
"This means that a height of 20 meters is achieved at 7 seconds in the restricted interval. In
fact, it turns out that h−1(t) = 7 · {π − arcsin[(t − 20)/18]}/π when h is restricted to the given
interval."
I'm not sure why my approach is incorrect; any pointers would be much appreciated!

Ryaners said:
I'm working through the problems in the Mooculus textbook as revision for Calculus I & there seems to be something wrong with how I'm manipulating the function to find its inverse in the following example.

## Homework Statement

The height in meters of a person off the ground as they ride a Ferris Wheel can be modeled by:
h(t) = 18 · sin[(π · t )/ 7] + 20
where t is time elapsed in seconds. If h is restricted to the domain [3.5, 10.5], find and interpret the meaning of h-1(20).

## The Attempt at a Solution

Here are my workings:

Let y = h(t)
y = 18⋅Sin[(π⋅t)/7] + 20
(y - 20) / 18 = Sin[(π⋅t) / 7]
arcsin [(y - 20) / 18] = (π⋅t) / 7
π⋅t = 7⋅{arcsin [(y - 20) / 18]}
t = (7/π)⋅{arcsin [(y - 20) / 18]}

⇒ h-1(t) = (7/π)⋅{arcsin [(t - 20) / 18]}
⇒ h-1(20) = (7/π)⋅arcsin(0) = 0

However this is the solution given:
h-1(20) = 7
"This means that a height of 20 meters is achieved at 7 seconds in the restricted interval. In
fact, it turns out that h−1(t) = 7 · {π − arcsin[(t − 20)/18]}/π when h is restricted to the given
interval."
I'm not sure why my approach is incorrect; any pointers would be much appreciated!
You forgot about the condition: "If h is restricted to the domain [3.5, 10.5]"
arcsin in a multivalued function, that makes calculating the inverse of h tricky.
(I mean ##\sin(0)=0##, but also ##\sin(\pi)=0##, etc...)

Maybe just set h(t) = 20, and see what value(s) in the domain [3.5, 10.5] satisfy this equation.

Ryaners said:
I'm working through the problems in the Mooculus textbook as revision for Calculus I & there seems to be something wrong with how I'm manipulating the function to find its inverse in the following example.

## Homework Statement

The height in meters of a person off the ground as they ride a Ferris Wheel can be modeled by:
h(t) = 18 · sin[(π · t )/ 7] + 20
where t is time elapsed in seconds. If h is restricted to the domain [3.5, 10.5], find and interpret the meaning of h-1(20).

## The Attempt at a Solution

Here are my workings:

Let y = h(t)
y = 18⋅Sin[(π⋅t)/7] + 20
(y - 20) / 18 = Sin[(π⋅t) / 7]
arcsin [(y - 20) / 18] = (π⋅t) / 7
π⋅t = 7⋅{arcsin [(y - 20) / 18]}
t = (7/π)⋅{arcsin [(y - 20) / 18]}

⇒ h-1(t) = (7/π)⋅{arcsin [(t - 20) / 18]}
⇒ h-1(20) = (7/π)⋅arcsin(0) = 0

However this is the solution given:
h-1(20) = 7
"This means that a height of 20 meters is achieved at 7 seconds in the restricted interval. In
fact, it turns out that h−1(t) = 7 · {π − arcsin[(t − 20)/18]}/π when h is restricted to the given
interval."
I'm not sure why my approach is incorrect; any pointers would be much appreciated!

If h is restricted to [3.5,10.5] (as you stated) then findig where h(t) = 20 makes no sense. Did you mean t in [3.5,10.5]? If that is what you mean then I would avoid starting with the arcsin function and just look at the equation h(t) = 20 directly; later, after clarifying the rough locations of the relevant root or roots, I would increase the accuracy by then using the arcsiin function at that point. Remember: the standard arcsin function yields angles between -π/2 and +π/2.

Ray Vickson said:
If h is restricted to [3.5,10.5] (as you stated) then findig where h(t) = 20 makes no sense. Did you mean t in [3.5,10.5]? If that is what you mean then I would avoid starting with the arcsin function and just look at the equation h(t) = 20 directly; later, after clarifying the rough locations of the relevant root or roots, I would increase the accuracy by then using the arcsiin function at that point. Remember: the standard arcsin function yields angles between -π/2 and +π/2.

Samy_A said:
Maybe just set h(t) = 20, and see what value(s) in the domain [3.5, 10.5] satisfy this equation.

Thanks Ray - it's asking me to find h-1(20), not h(20). Apologies if I've misinterpreted your reply or that was a typo. The question is exactly as stated in the textbook & I'm disinclined to think it's a mistake. It's pretty confusing, though.
If h(t) has an inverse h-1(t) (on the interval h ∈ [3.5, 10.5]), the 'output' of h(20) will be the height h at t=20, right? Is h-1(20) not then the time at h = 20, which is not a permitted value of h? Maybe it is a mistake, and they do mean t ∈ [3.5, 10.5]. (Those seem like more sensible as values for time than for height with a ferris wheel in mind, though maybe bringing the real world into textbook examples is a bad idea..!)

It says: "If h is restricted to the domain [3.5, 10.5]", so they quite clearly mean that you have to find a t ∈ [3.5, 10.5] satisfying h(t)=20.

Samy_A said:
It says: "If h is restricted to the domain [3.5, 10.5]", so they quite clearly mean that you have to find a t ∈ [3.5, 10.5] satisfying h(t)=20.
Ah yes ok, h being a function & not a variable. My bad. Thanks for pointing that out.

Ryaners said:
Thanks Ray - it's asking me to find h-1(20), not h(20). Apologies if I've misinterpreted your reply or that was a typo. The question is exactly as stated in the textbook & I'm disinclined to think it's a mistake. It's pretty confusing, though.
If h(t) has an inverse h-1(t) (on the interval h ∈ [3.5, 10.5]), the 'output' of h(20) will be the height h at t=20, right? Is h-1(20) not then the time at h = 20, which is not a permitted value of h? Maybe it is a mistake, and they do mean t ∈ [3.5, 10.5]. (Those seem like more sensible as values for time than for height with a ferris wheel in mind, though maybe bringing the real world into textbook examples is a bad idea..!)

No, I knew perfectly well what you MEANT, which is different from what you SAID. Here it is exactly (via cut and paste): "If h is restricted to the domain [3.5, 10.5], find and interpret the meaning of h-1(20)". As written, that statement makes no sense; ##h^{-1}(20)## refers to the solution of the equation ##h(t) = 20##, but you said you were restricting ##h## to [3.5, 10.5], which most definitely excludes the value '20' for h. Beginners such as students and self-learners might have more trouble figuring what you actually meant, so that is why it is always good to avoid such errors if you can. In this case it seems that your textbook needs some editing, so the problem begins there.

I suppose one could say that the word "domain" clears it up, because the domain of h(t) is a set of "allowed" t-values. However, it is a somewhat weird way of saying it, and as I indicated, beginners might have trouble interpreting it. It would have been better if they said "the domain of h is [3.5,10.5]" or something similar.

Last edited:

## 1. What is an inverse of a Sin function?

The inverse of a Sin function is a function that "undoes" the action of the Sin function. It takes the output of the Sin function and gives the input value that produced that output.

## 2. Why is finding the inverse of a Sin function important?

Finding the inverse of a Sin function is important because it allows us to solve equations involving Sin functions. It also helps us understand and analyze the behavior of Sin functions.

## 3. How do you find the inverse of a Sin function?

To find the inverse of a Sin function, you first need to switch the x and y variables. Then, solve for y and replace it with f^-1(x). Lastly, use inverse trigonometric identities to simplify the equation.

## 4. What are some common mistakes when finding the inverse of a Sin function?

One common mistake is forgetting to switch the x and y variables. Another mistake is not using the correct inverse trigonometric identities to simplify the equation. It is also important to check for extraneous solutions when solving for the inverse.

## 5. Can the inverse of a Sin function always be found?

No, the inverse of a Sin function can only be found if the Sin function is one-to-one, meaning it passes the horizontal line test. If the Sin function is not one-to-one, it does not have an inverse.

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