# Inverse trigonometric equation

by utkarshakash
Tags: equation, inverse, trigonometric
 PF Gold P: 824 1. The problem statement, all variables and given/known data Solve $cot^{-1} x - cot^{-1} (n^2 - x+1)=cot^-1(n-1)$ 2. Relevant equations 3. 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 $(n^2+1-2x)(n-1) = x(n^2-x+1)$
 P: 154 Are you asking how to solve that equation?
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P: 7,800
 Quote by utkarshakash 1. The problem statement, all variables and given/known data Solve $cot^{-1} x - cot^{-1} (n^2 - x+1)=cot^-1(n-1)$ 2. Relevant equations 3. 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:

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

PF Gold
P: 824
Inverse trigonometric equation

 Quote by MrWarlock616 Are you asking how to solve that equation?
Yes I have to find the value of x
PF Gold
P: 824
 Quote by SammyS 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?
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P: 7,800
 Quote by utkarshakash But what should be the next step?
Multiply each side out.

Get everything to the left side.

You have a quadratic equation in x.
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P: 824
 Quote by SammyS Multiply each side out. Get everything to the left side. You have a quadratic equation in x.
But its discriminant is way too big!
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P: 7,800
 Quote by utkarshakash But its discriminant is way too big!
Too big in what sense ?

What do you get for the discriminant ?
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P: 824
 Quote by SammyS 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=$n^4+6n^2-8n+2$
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P: 7,800
 Quote by utkarshakash 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?
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P: 824
 Quote by SammyS 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
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PF Gold
P: 7,800
 Quote by utkarshakash 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.
 Emeritus Sci Advisor HW Helper PF Gold P: 7,800 I still get a quadratic equation in x, so that can't be the solution. $\displaystyle \cot(A+B)=\frac{\cot(A)\,\cot(B)-1}{\cot(A)+\cot(B)}$
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P: 824
 Quote by SammyS 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

$-4n^3+6n^2-8n+9$
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PF Gold
P: 7,800
 Quote by utkarshakash 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.
PF Gold
P: 824
 Quote by SammyS 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?
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Thanks
P: 9,650
 Quote by utkarshakash 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.
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 Quote by haruspex 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?

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