# Equation involving inverse trigonometric function

• phymath7
In summary, taking the sine of both sides of the equation yields a solution that is within the red circle.
phymath7
Homework Statement
Solve the following equation:
$$\sin^{-1}(k)=\frac {k^2}{2} +1$$ where 0<k<1.
Relevant Equations
Maclaurin series expansion for ##sin^{-1}(k)##
I came across the mentioned equation aftet doing a integral for an area related problem.Doing the maclaurin series expansion for the inverse sine function,I considered the first two terms(as the latter terms involved higher power of the argument divided by factorial of higher numbers),doing so the equation then became like the following:
$$k+\frac {k^3}{6}=\frac{k^2}{2} +1$$
But this equation gives one real solution and two complex solution.And the real solution is greater than 1 which doesn't satisfies the condition.So what's wrong?Do I need to consider more terms in the series?But doing so makes the equation much more hardr to solve which I guess the author hasn't intended in the problem.

For x close to 1 your lefthand side does not exceed the 1.5 from the righthand side. But arcsin does (##\pi/2##), so your approach doesn't work...

BvU said:
For x close to 1 your lefthand side does not exceed the 1.5 from the righthand side. But arcsin does (##\pi/2##), so your approach doesn't work...
What is x?The equation doesn't contain x.Moreover,k can't exceed 1,so your assumption(taking k as (##\pi/2##))is wrong.Perhaps you didn't notice it's an inverse trig function.

Last edited:
phymath7 said:
Homework Statement: Solve the following equation:
$$\sin^{-1}(k)=\frac {k^2}{2} +1$$ where 0<k<1.
Relevant Equations: Maclaurin series expansion for ##sin^{-1}(k)##

I came across the mentioned equation aftet doing a integral for an area related problem. Doing the maclaurin series expansion for the inverse sine function,I considered the first two terms(as the latter terms involved higher power of the argument divided by factorial of higher numbers),doing so the equation then became like the following:
$$k+\frac {k^3}{6}=\frac{k^2}{2} +1$$
The left side is not the Maclaurin series expansion for the inverse sine function. It's the first couple of nonzero terms for the expansion of the sine function.
phymath7 said:
But this equation gives one real solution and two complex solution.And the real solution is greater than 1 which doesn't satisfies the condition.So what's wrong?Do I need to consider more terms in the series?But doing so makes the equation much more hardr to solve which I guess the author hasn't intended in the problem.
Or maybe your setup for the area calculation was incorrect. The equation you're asked to solve doesn't seem to have a solution other than by numerical means, so I would suggest going back to the original problem to see if you've made a mistake there.

sin(1.5)=.997
I would take the sine of both sides and do a taylor series around 1.

SammyS
Mark44 said:
The left side is not the Maclaurin series expansion for the inverse sine function. It's the first couple of nonzero terms for the expansion of the sine function.
Or maybe your setup for the area calculation was incorrect. The equation you're asked to solve doesn't seem to have a solution other than by numerical means, so I would suggest going back to the original problem to see if you've made a mistake there.
The left side is the maclaurin series for inverse sine function.Have a look in here:https://mathworld.wolfram.com/MaclaurinSeries.html.

Mark44 said:
Or maybe your setup for the area calculation was incorrect. The equation you're asked to solve doesn't seem to have a solution other than by numerical means, so I would suggest going back to the original problem to see if you've made a mistake there.
There's no error in the setup(checked from the solution manual)

Frabjous said:
sin(1.5)=.997
Where does this assumption come from?How does that help?
Frabjous said:
I would take the sine of both sides and do a taylor series around 1.
Taking sine of both sides leave us with:
$$k=sin( \frac {k^2}{2} +1)$$.How do we proceed?

phymath7 said:
The left side is the maclaurin series for inverse sine function.Have a look in here:https://mathworld.wolfram.com/MaclaurinSeries.html.
You're right. The first couple of terms of the Maclaurin series for sin(x) looks similar -- ##x - \frac {x^3}6 \pm \dots##, and that fooled me into thinking that you had used the series for sine rather than arcsine.

phymath7 said:
What is x?The equation doesn't contain x.Moreover,k can't exceed 1,so your assumption(taking k as (##\pi/2##))is wrong.Perhaps you didn't notice it's an inverse trig function.
Sorry, old habit to use ##x## as independent variable on horizontal axis. Your ##k##, to be sure.
And I did not say ##k = \pi/2## but the value of the ##\arcsin## function near ##k=1##.A simple plot clarifies things. Of course, you had made it already ?

Clearly the solution is within the red circle. So ##k## should be close to 1, and any series expansion should be around 1, not around zero.

##\ ##

phymath7 said:
Where does this assumption come from?How does that help?

Taking sine of both sides leave us with:
$$k=sin( \frac {k^2}{2} +1)$$.How do we proceed?
I then plugged in the bounds, 0 and 1. Since k=1 is close to the solution, it is preferable to take series expansion there because it will require less terms for the given number of significant digits required.

Frabjous said:
I then plugged in the bounds, 0 and 1. Since k=1 is close to the solution, it is preferable to take series expansion there because it will require less terms for a given number of significant digits required.
It' clear from graphing utility that k is close to 1,but how do we use that guess?Mainly,I am asking how my Maclaurin series expansion is wrong?Do I need to take more terms?If I do ,then things get more complicated(due to terms with higher power).How to overcome this?

phymath7 said:
It' clear from graphing utility that k is close to 1,but how do we use that guess?Mainly,I am asking how my Maclaurin series expansion is wrong?Do I need to take more terms?If I do ,then things get more complicated(due to terms with higher power).How to overcome this?
Why are you wedded to the Maclaurin series? It is preferrable to do series expansions as close as possible to the solution so that you can minimize the number of terms that you need.

Frabjous said:
It is preferrable to do series expansions as close as possible to the solution so that you can minimize the number of terms that you need.
Do I take a=1 in the Taylor series to do the series expansion?That's what you mean?

Yes. I take it you can accept a numerical solution ?

##\ ##

phymath7 said:
Do I take a=1 in the Taylor series to do the series expansion?That's what you mean?
Yes. You are expanding in powers of k-a. For a=0 (maclaurin) you are doing powers near 1. For a=1 you are doing powers near 0 which has much better convergence.

phymath7
There's a small problem: ##f'(1)## doesn't exist for ##f=\arcsin##.
A numerical solution, then ?

BvU said:
There's a small problem: ##f'(1)## doesn't exist for ##f=\arcsin##.
A numerical solution, then ?

View attachment 324618
Which is why I said to take the sine of both sides of the equation in post #5.

Works like a dream !

SammyS and Frabjous

## 1. What is an inverse trigonometric function?

An inverse trigonometric function is a mathematical function that takes the output of a trigonometric function (such as sine, cosine, or tangent) and finds the input angle that produced that output. It essentially "undoes" the trigonometric function.

## 2. What is the difference between a trigonometric function and an inverse trigonometric function?

A trigonometric function takes an angle as its input and gives a ratio or value as its output. An inverse trigonometric function takes a ratio or value as its input and gives the corresponding angle as its output.

## 3. What is the notation used for inverse trigonometric functions?

The notation used for inverse trigonometric functions is "sin-1(x)" for arcsine, "cos-1(x)" for arccosine, and "tan-1(x)" for arctangent. This notation is read as "the inverse sine/cosine/tangent of x."

## 4. How do you solve an equation involving inverse trigonometric functions?

To solve an equation involving inverse trigonometric functions, you first isolate the inverse trigonometric function on one side of the equation. Then, use the corresponding trigonometric function to both sides of the equation to "undo" the inverse trigonometric function. Finally, solve for the variable using algebraic techniques.

## 5. What are some common applications of inverse trigonometric functions?

Inverse trigonometric functions are commonly used in fields such as physics, engineering, and navigation to solve problems involving angles and distances. They are also used in computer graphics and animation to create realistic movements and rotations.

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