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Two related questions, one easy, one just a bit harder:

1) Prove the identity $$ \tan^{-1} \left( \frac{x}{1-x^2} \right) = \tan^{-1}x + \tan^{-1}(x^3) $$

2) Now try to find a geometric or trigonometric proof of that same geometry, without resorting to calculus.

I'll post an answer to both questions in a couple of days if nobody has one yet.

symplectomorphic
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Mark Fischler
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    for 1), just take the tangent of both sides, and use the sum formula for tangent... – symplectomorphic May 28 '14 at 04:28
  • I agree with the above comment, it should count as a trignonometric proof to use the tangent sum formula. – user111187 May 28 '14 at 04:38
  • let me be clear: I don't mean that one should assume the desired identity, and then take the tangent of both sides in order to deduce it, which would be circular. my comment was meant as a hint that the sum formula tangent will be helpful... (as heropup has now spelled out). – symplectomorphic May 28 '14 at 04:57
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    of related interest: http://math.stackexchange.com/questions/326334/a-question-about-the-arctangent-addition-formula – symplectomorphic May 28 '14 at 05:23

3 Answers3

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The proof can be established starting from the angle addition identity $$\sin(x+y) = \sin x \cos y + \cos x \sin y.$$ Then with $x = \frac{\pi}{2} - a$, $y = -b$, we get $$\cos(a+b) = \sin(\tfrac{\pi}{2} - a - b) = \cos a \cos b - \sin a \sin b.$$ We then find $$\tan(x+y) = \frac{\sin(x+y)}{\cos(x+y)} = \frac{(\sin x \cos y + \cos x \sin y)/(\cos x \cos y)}{(\cos x \cos y - \sin x \sin y)/(\cos x \cos y)} = \frac{\tan x + \tan y}{1 - \tan x \tan y}.$$ Now letting $u = \tan x$, $v = \tan y$, we get $$\tan(\tan^{-1} u + \tan^{-1}v) = \frac{u+v}{1-uv},$$ or $$\tan^{-1} u + \tan^{-1} v = \tan^{-1} \frac{u+v}{1-uv}.$$ Now letting $u = x$, $v= x^3$, we find $$\frac{u+v}{1-uv} = \frac{x+x^3}{1-x^4} = \frac{x(1+x^2)}{(1-x^2)(1+x^2)} = \frac{x}{1-x^2},$$ and the result immediately follows.

heropup
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  • just to be clear, your deduction indicated by "or" (where you've just taken the inverse tangent of both sides) relies on the fact that $\tan^{-1}(\tan a)=a$, which is true only if $a\in(-\pi/2,\pi/2)$. in particular, you're assuming $\tan^{-1}u+\tan^{-1}v=\tan^{-1}(\tan x)+\tan^{-1}(\tan y)\in(-\pi/2,\pi/2)$. – symplectomorphic May 28 '14 at 05:11
  • I am well aware of that. There is no issue when we regard the inverse tangent as a multivalued mapping for the purposes of establishing the identity as it was given in the question. In other words, if you want to be pedantic, the given identity is not strictly true for all choices of $x$ if you take the codomain of $\tan^{-1}$ to be $(-\pi/2, \pi/2)$. But since that was left out, I saw no reason to be so picky. The identity as stated works fine for equating sets to sets (or equivalence classes, if you prefer to think of it that way). – heropup May 28 '14 at 05:24
  • I wasn't criticizing your answer (I upvoted it); I was just making the assumption explicit for students who want to be totally rigorous. – symplectomorphic May 28 '14 at 05:27
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Part I: Consider $f(x) = tan^{-1}\left(\dfrac{x}{1-x^2}\right) - tan^{-1}x - tan^{-1}(x^3)$, taking derivative of $f$:

$f'(x) = \dfrac{1+x^2}{1-x^2+x^4} - \dfrac{1}{1+x^2} - \dfrac{3x^2}{1+x^6} = \dfrac{(1+x^2)^2 - (1-x^2+x^4) - 3x^2}{1+x^6} = \dfrac{1+ 2x^2 + x^4 - 1 + x^2 - x^4 - 3x^2}{1+x^6} = 0$. Thus $f$ is constant on $\mathbb{R}$, and $f(x) = f(0) = 0$, and the identity follows.

DeepSea
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Part I)

Taking the tan of the RHS, we have a form $tan(A+B)$ where $A = tan^{-1}(x)$, $B = tan^{-1}(x^3)$. $$ tan(A+B) = \frac{tan(A)+tan(B)}{1-tan(A)*tan(B)} = \frac{x + x^3}{1 - x^4} = \frac{x(1 + x^2)}{(1-x^2)(1+x^2)} = \frac{x}{1-x^2} $$

Now the tan of LHS is also $\frac{x}{1-x^2}$.

Therefore the identity is true.

Orange Peel
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  • this interprets my comment too literally. you've assumed $a=b$ and then derived $\tan a=\tan b$, when the goal was to show $a=b$. my comment was meant more as a hint, not a direct proof strategy. – symplectomorphic May 28 '14 at 04:55
  • But if tan a = tan b, then a = b (with restrictions to quadrants) – Orange Peel May 28 '14 at 04:59
  • yes... but as written, your proof is circular, and you haven't explained how to deduce $a=b$ from $\tan a=\tan b$. – symplectomorphic May 28 '14 at 05:04
  • Do you mean my second last line? I derived the tan of LHS by taking literally taking the tan of the LHS, not just by using the result from the RHS. – Orange Peel May 28 '14 at 05:10
  • I think it would have been better not to refer to RHS and LHS, which makes it sound as if you're taking the given equation for granted, when you're supposed to be proving it. in other words, it would've been better to write: by the sum formula for tangent, we have $\tan(\tan^{-1}x+\tan^{-1}(x^3))=\frac{x+x^3}{1-x^4}=\frac{x}{1-x^2}$. we also have $\tan(\tan^{-1}(\frac{x}{1-x^2}))=\frac{x}{1-x^2}$. therefore $\tan(\tan^{-1}x+\tan^{-1}(x^3))=\tan(\tan^{-1}(\frac{x}{1-x^2}))$. then you can make a comment about deducing $a=b$ from $\tan a=\tan b$. – symplectomorphic May 28 '14 at 05:16
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    It is perfectly legitimate to refer to the LHS and the RHS in forming a proof, but I see where you're coming from. – Orange Peel May 28 '14 at 05:23
  • I didn't say it wasn't legitimate: I said it is misleading. the logic of your answer is not at all obvious. you leave a lot to be desired when you write "therefore the identity is true." – symplectomorphic May 28 '14 at 05:25