# Inverse trigonometric equation

• utkarshakash
In summary: If you can't solve it, you might want to try a different method. So as haruspex said, it's a messy...quadratic equation in x. If you can't solve it, you might want to try a different method.
utkarshakash
Gold Member

## Homework Statement

Solve

$cot^{-1} x - cot^{-1} (n^2 - x+1)=cot^-1(n-1)$

## The Attempt at a Solution

I can write the above equation as θ+α=β where the symbols represent the respective inverse functions. Now I take tan of both sides. Simplifying I get

$(n^2+1-2x)(n-1) = x(n^2-x+1)$

Are you asking how to solve that equation?

utkarshakash said:

## Homework Statement

Solve

$cot^{-1} x - cot^{-1} (n^2 - x+1)=cot^-1(n-1)$

## The Attempt at a Solution

I can write the above equation as θ+α=β where the symbols represent the respective inverse functions. Now I take tan of both sides. Simplifying I get

$(n^2+1-2x)(n-1) = x(n^2-x+1)$
I get nearly the same thing:

$(n^2+1-2x)(n-1) = 1+x(n^2-x+1)$

MrWarlock616 said:
Are you asking how to solve that equation?

Yes I have to find the value of x

SammyS said:
I get nearly the same thing:

$(n^2+1-2x)(n-1) = 1+x(n^2-x+1)$

But what should be the next step?

utkarshakash said:
But what should be the next step?
Multiply each side out.

Get everything to the left side.

You have a quadratic equation in x.

SammyS said:
Multiply each side out.

Get everything to the left side.

You have a quadratic equation in x.

But its discriminant is way too big!

utkarshakash said:
But its discriminant is way too big!

Too big in what sense ?

What do you get for the discriminant ?

SammyS said:
Too big in what sense ?

What do you get for the discriminant ?

How are you getting that 'extra' 1 in the RHS. OK I'm only posting what I got from my answer i.e. without that '1'.

D=$n^4+6n^2-8n+2$

utkarshakash said:
How are you getting that 'extra' 1 in the RHS. OK I'm only posting what I got from my answer i.e. without that '1'.
...]
Wow! I showed that "extra" 1 way back in post #3. Now you mention it?

I got it because,

$\displaystyle \left(1+\frac{1}{x(n^2-x+1)}\right)x(n^2-x+1)=x(n^2-x+1)+1\ .$

How did you not get it?

SammyS said:
Wow! I showed that "extra" 1 way back in post #3. Now you mention it?

I got it because,

$\displaystyle \left(1+\frac{1}{x(n^2-x+1)}\right)x(n^2-x+1)=x(n^2-x+1)+1\ .$

How did you not get it?

Hey there is a much better solution than this. Instead of taking tan if I take cot of both sides I get a very simple result and here's it

$x= \dfrac{n(n+1)}{2}$

And yes you said right. There was a calculation error in my solution. It should have that extra 1

utkarshakash said:
Hey there is a much better solution than this. Instead of taking tan if I take cot of both sides I get a very simple result and here's it

$x= \dfrac{n(n+1)}{2}$
...

That is interesting. I'll have to take a look at it.

I still get a quadratic equation in x, so that can't be the solution.

$\displaystyle \cot(A+B)=\frac{\cot(A)\,\cot(B)-1}{\cot(A)+\cot(B)}$

SammyS said:
I still get a quadratic equation in x, so that can't be the solution.

$\displaystyle \cot(A+B)=\frac{\cot(A)\,\cot(B)-1}{\cot(A)+\cot(B)}$

OMG I made the silliest mistake ever. Instead of taking cot of both sides I just removed the 'inverse thing' and simply calculated the value of x from there. OK sticking to my previous solution I take tan of both sides and get the same thing what you got(with that extra 1) and get the discriminant as

$-4n^3+6n^2-8n+9$

utkarshakash said:
OMG I made the silliest mistake ever. Instead of taking cot of both sides I just removed the 'inverse thing' and simply calculated the value of x from there. OK sticking to my previous solution I take tan of both sides and get the same thing what you got(with that extra 1) and get the discriminant as

$-4n^3+6n^2-8n+9$
What I got for a discriminant was

$n^4+6n^2-8n+9\ .$

I just rechecked it using WolframAlpha on

$(n^2+1-2x)(n-1) = 1+x(n^2-x+1)$

and it confirmed my result.

SammyS said:
What I got for a discriminant was

$n^4+6n^2-8n+9\ .$

I just rechecked it using WolframAlpha on

$(n^2+1-2x)(n-1) = 1+x(n^2-x+1)$

and it confirmed my result.

OK so what should I do after getting discriminant?

utkarshakash said:
OK so what should I do after getting discriminant?
Nothing special. It's just a quadratic equation in x; solve it with the usual formula. If it's messy, it's messy.

haruspex said:
Nothing special. It's just a quadratic equation in x; solve it with the usual formula. If it's messy, it's messy.

You can't simply say that I cannot get rid of that square root. Please help me. Its not even a perfect square. Should I leave it this way?

utkarshakash said:
You can't simply say that I cannot get rid of that square root. Please help me. Its not even a perfect square. Should I leave it this way?
If n=0, then the discriminant is a perfect square. I found no other cases. Also, the discriminant is positive for all values of n.

So as haruspex said, it's a messy result.

SammyS said:
If n=0, then the discriminant is a perfect square. I found no other cases. Also, the discriminant is positive for all values of n.

So as haruspex said, it's a messy result.

OK so you want me to continue with that square root.

utkarshakash said:
OK so you want me to continue with that square root.

That's about all there is to do, but you can't do much simplifying.

SammyS said:
That's about all there is to do, but you can't do much simplifying.

OK Thanks!

## What is an "inverse trigonometric equation"?

An inverse trigonometric equation is a mathematical equation that involves finding the measure of an angle in a triangle using the inverse trigonometric functions: sine, cosine, tangent, cosecant, secant, and cotangent.

## What are the inverse trigonometric functions?

The inverse trigonometric functions are arcsine, arccosine, arctangent, arccosecant, arcsecant, and arccotangent. These functions are used to find the measure of an angle in a triangle when given the ratio of sides.

## How do I solve an inverse trigonometric equation?

To solve an inverse trigonometric equation, you must first isolate the trigonometric function on one side of the equation. Then, use the inverse trigonometric function on both sides to find the measure of the angle.

## What is the domain and range of inverse trigonometric functions?

The domain of inverse trigonometric functions is restricted to certain values depending on the specific function. The range of inverse trigonometric functions is typically between -π/2 and π/2 for arcsine, -π/2 and π/2 for arccosine, and -π/2 and π/2 for arctangent.

## How are inverse trigonometric functions used in real life?

Inverse trigonometric functions are used in real life to solve real-world problems involving angles and distances. For example, they can be used in navigation and surveying, as well as in physics and engineering for calculating forces and trajectories.

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