Inverse trigonometric equation


by utkarshakash
Tags: equation, inverse, trigonometric
utkarshakash
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#1
Jan5-13, 10:07 AM
P: 658
1. The problem statement, all variables and given/known data
Solve

[itex]cot^{-1} x - cot^{-1} (n^2 - x+1)=cot^-1(n-1)[/itex]

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

[itex](n^2+1-2x)(n-1) = x(n^2-x+1)[/itex]
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MrWarlock616
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#2
Jan5-13, 01:25 PM
P: 154
Are you asking how to solve that equation?
SammyS
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#3
Jan5-13, 01:43 PM
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Quote Quote by utkarshakash View Post
1. The problem statement, all variables and given/known data
Solve

[itex]cot^{-1} x - cot^{-1} (n^2 - x+1)=cot^-1(n-1)[/itex]

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

[itex](n^2+1-2x)(n-1) = x(n^2-x+1)[/itex]
I get nearly the same thing:

[itex](n^2+1-2x)(n-1) = 1+x(n^2-x+1)[/itex]

utkarshakash
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#4
Jan6-13, 12:53 AM
P: 658

Inverse trigonometric equation


Quote Quote by MrWarlock616 View Post
Are you asking how to solve that equation?
Yes I have to find the value of x
utkarshakash
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#5
Jan6-13, 12:53 AM
P: 658
Quote Quote by SammyS View Post
I get nearly the same thing:

[itex](n^2+1-2x)(n-1) = 1+x(n^2-x+1)[/itex]
But what should be the next step?
SammyS
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#6
Jan6-13, 01:36 AM
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Quote Quote by utkarshakash View Post
But what should be the next step?
Multiply each side out.

Get everything to the left side.

You have a quadratic equation in x.
utkarshakash
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#7
Jan6-13, 08:54 AM
P: 658
Quote Quote by SammyS View Post
Multiply each side out.

Get everything to the left side.

You have a quadratic equation in x.
But its discriminant is way too big!
SammyS
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#8
Jan6-13, 02:43 PM
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Quote Quote by utkarshakash View Post
But its discriminant is way too big!
Too big in what sense ?

What do you get for the discriminant ?
utkarshakash
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#9
Jan7-13, 02:32 AM
P: 658
Quote Quote by SammyS View Post
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=[itex]n^4+6n^2-8n+2[/itex]
SammyS
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#10
Jan7-13, 08:42 AM
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Quote Quote by utkarshakash View Post
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,

[itex]\displaystyle \left(1+\frac{1}{x(n^2-x+1)}\right)x(n^2-x+1)=x(n^2-x+1)+1\ .[/itex]

How did you not get it?
utkarshakash
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#11
Jan8-13, 06:23 AM
P: 658
Quote Quote by SammyS View Post
Wow! I showed that "extra" 1 way back in post #3. Now you mention it?

I got it because,

[itex]\displaystyle \left(1+\frac{1}{x(n^2-x+1)}\right)x(n^2-x+1)=x(n^2-x+1)+1\ .[/itex]

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

[itex] x= \dfrac{n(n+1)}{2} [/itex]

And yes you said right. There was a calculation error in my solution. It should have that extra 1
SammyS
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#12
Jan8-13, 01:53 PM
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Quote Quote by utkarshakash View Post
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

[itex] x= \dfrac{n(n+1)}{2} [/itex]
...
That is interesting. I'll have to take a look at it.
SammyS
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#13
Jan9-13, 01:41 AM
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I still get a quadratic equation in x, so that can't be the solution.

[itex]\displaystyle \cot(A+B)=\frac{\cot(A)\,\cot(B)-1}{\cot(A)+\cot(B)}[/itex]
utkarshakash
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#14
Jan9-13, 06:34 AM
P: 658
Quote Quote by SammyS View Post
I still get a quadratic equation in x, so that can't be the solution.

[itex]\displaystyle \cot(A+B)=\frac{\cot(A)\,\cot(B)-1}{\cot(A)+\cot(B)}[/itex]
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

[itex] -4n^3+6n^2-8n+9[/itex]
SammyS
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#15
Jan9-13, 12:22 PM
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Quote Quote by utkarshakash View Post
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

[itex] -4n^3+6n^2-8n+9[/itex]
What I got for a discriminant was

[itex]n^4+6n^2-8n+9\ .[/itex]

I just rechecked it using WolframAlpha on

[itex](n^2+1-2x)(n-1) = 1+x(n^2-x+1)[/itex]

and it confirmed my result.
utkarshakash
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#16
Jan9-13, 12:51 PM
P: 658
Quote Quote by SammyS View Post
What I got for a discriminant was

[itex]n^4+6n^2-8n+9\ .[/itex]

I just rechecked it using WolframAlpha on

[itex](n^2+1-2x)(n-1) = 1+x(n^2-x+1)[/itex]

and it confirmed my result.
OK so what should I do after getting discriminant?
haruspex
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#17
Jan9-13, 02:53 PM
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Quote Quote by utkarshakash View Post
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.
utkarshakash
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#18
Jan10-13, 10:05 AM
P: 658
Quote Quote by haruspex View Post
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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