Finding the Inverse Function of a Trigonometric Expression

[mtayab1994]

Homework Statement

There were other questions before this one but i solved them all.

Find the inverse function of [tex]f(x)=arctan(\sqrt{1+x^{2}}-x)[/tex] for every x in the interval ]0,pi/2[ .That's the interval that I found when counting f(R) because f is a bijection from R to f(R). Hence f(R)=J=]0,pi/2[. ( just in case someone was wondering where i got it from?

The Attempt at a Solution

For every x in J we get :

[tex]
\begin{eqnarray*}
x=arctan(\sqrt{1+y^{2}}-y)\Rightarrow tan(x)=tan(arctan(\sqrt{1+y^{2}}-y) & \Rightarrow & tan(x)=\sqrt{1+y^{2}}-y\\
& \Rightarrow & tan(x)+y=\sqrt{1+y^{2}}\\
& \Rightarrow & tan^{2}(x)+2ytan(x)+y^{2}=1+y^{2}
\end{eqnarray*}[/tex]

then [tex]tan^{2}(x)+2ytan(x)+y^{2}=1+y^{2}\Rightarrow tan(x)[tan(x)+2y]=1\Rightarrow tan(x)+2y=\frac{1}{tan(x)}\Rightarrow tan(x)+2y=tan(\frac{\pi}{2}-x)[/tex]

then [tex]tan(x)+2y=tan(\frac{\pi}{2}-x)\Longrightarrow2y=tan(\frac{\pi}{2}-2x)\Rightarrow y=\frac{1}{2}[tan(\frac{\pi}{2}-2x)][/tex]

Is that correct, because the given answer here is: [tex]f^{-1}(x)=tan(\frac{\pi}{2}-2x)[/tex]

 

[micromass]

then [tex]tan(x)+2y=tan(\frac{\pi}{2}-x)\Longrightarrow2y=tan(\frac{\pi}{2}-2x)[/tex]

I'm not really seeing how you did this step.

 

[mtayab1994]

I'm not really seeing how you did this step.

I did 2y= tan(pi/2-x)-tan(x) and i did 2y=tan(pi/2-x-x) and then i got 2y=tan(pi/2-2x)

 

[micromass]

And I don't see why

[tex]\tan(\pi/2 - x)-\tan(x)=\tan(\pi/2 - x-x)[/tex]

 

[mtayab1994]

And I don't see why

[tex]\tan(\pi/2 - x)-\tan(x)=\tan(\pi/2 - x-x)[/tex]

Why not??

 

[micromass]

Why not??

If you think the equality holds, then it's up to you to prove it. I just don't see where it comes form. Do you have a justification for the equality?
If you do not find a justification, then you might think that it is not true. Try some easy values for x and see if the equality holds. If you found one value for which it does not hold, then you have disproven the equality.

 

[mtayab1994]

If you think the equality holds, then it's up to you to prove it. I just don't see where it comes form. Do you have a justification for the equality?
If you do not find a justification, then you might think that it is not true. Try some easy values for x and see if the equality holds. If you found one value for which it does not hold, then you have disproven the equality.

I'm sorry I was lost when thinking about that. It should be [tex]y=\frac{1}{2}[tan(\frac{\pi}{2}-x)-tan(x)][/tex] but then when you want to simplify what can you do??

 

[micromass]

You need to prove somehow that

[tex]2\tan(\pi/2 -2x)=\tan(\pi/2 -x) - \tan(x)[/tex]

So you see formula's that allow you to do that??

 

[SammyS]

... then [tex]\tan^{2}(x)+2y\tan(x)+y^{2}=1+y^{2}\ \Rightarrow\dots[/tex]

The y²'s cancel, leaving

[itex]\tan^{2}(x)+2y\tan(x)=1[/itex]

Solve that for y.

 

[mtayab1994]

The y²'s cancel, leaving

[itex]\tan^{2}(x)+2y\tan(x)=1[/itex]

Solve that for y.

y=(1-tan^2(x))/(2tan(x))

 

[SammyS]

y=(1-tan^2(x))/(2tan(x))

Right, and that's also the reciprocal of the identity for tan(2x) .