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I came across this problem to solve for $x$:

$$\sin[\cot^{-1}(x + 1)]=\cos[\tan^{-1}x]$$

I tried to do it initially by converting the cosine term on the right to sine by using $\sin(\frac{\pi}{2}-x)=\cos x$, but for some reason this doesn't work:

$$\sin[\cot^{-1}(x + 1)]=\sin[\frac{\pi}{2}-\tan^{-1}x]$$

$$\implies\cot^{-1}(x + 1)=\frac{\pi}{2}-\tan^{-1}x$$

$$\implies\cot^{-1}(x + 1)=\cot^{-1}x$$

$$\implies x + 1=x$$

$$\implies 1=0$$

Yet, if instead I use $\sin(\frac{\pi}{2}+x)=\cos x$:

$$\sin[\cot^{-1}(x + 1)]=\sin[\frac{\pi}{2}+\tan^{-1}x]$$

$$\implies\cot^{-1}(x + 1)=\frac{\pi}{2}+\tan^{-1}x$$

$$\implies\frac{\pi}{2}-\tan^{-1}(x + 1)=\frac{\pi}{2}+\tan^{-1}x$$

$$\implies\tan^{-1}(-x - 1)=\tan^{-1}x$$

$$\implies-x-1=x$$

$$\implies x=-\frac{1}{2}$$

And this is indeed the correct answer according to the book. What I don't get is why the first method failed. The identity $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$ should hold for all real numbers, so I don't think that is the problem. Moreover, if I plug $\frac 1 2$ and $-\frac 1 2$ into $\cot^{-1}x$, I get different answers, so this isn't a problem with the cancellation either. What's going wrong here?

EDIT: It appears that the problem is that since the value of $x$ is negative, the term $\frac \pi 2 - \tan^{-1}x$ becomes positive and lands outside the range where $\sin x$ is injective, and so the two inputs are different despite the fact that the output of their sine is equivalent. I guess a follow up question here is how one would figure this out without knowing that the answer is negative. It seems impossible to know which conversion to use beforehand without that knowledge.

Blue
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OmG
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  • Set $\dfrac1x=y$ $$\lim_{x\to\infty}\dfrac{x+1}x\to1$$ – lab bhattacharjee Sep 03 '19 at 12:41
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    Note that $\sin x=\sin y$ is not equivalent to $x=y$ – Vasili Sep 03 '19 at 12:41
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    Hi @Iab-bhattacharjee I don't think I understand what you're trying to say, could you explain a bit please? – OmG Sep 03 '19 at 13:07
  • I think this is a genuinely interesting problem. Kudos for spotting this. I like it! – Allawonder Sep 03 '19 at 14:00
  • @Vasya That's true, but that's not the problem here, for if you do use the correct argument in the first method, the $x,$ still disappears, leaving us with infinitely more contradictions. :) – Allawonder Sep 03 '19 at 14:01
  • @Allawonder I don't see what you mean about the "correct argument". Noting that if $\sin(\alpha) = \sin(\beta)$ then $\alpha = \beta + 2n\pi$ or $\alpha = (\pi - \beta) + 2n\pi$ (as in one of the answers below), $\pi - (\pi/2 - \arctan x) = \pi/2 + \arctan x$ and you get the same result from that branch of the first method as you do from the second method. – David K Jun 16 '22 at 13:27
  • The error isn't in the step where you convert cos to sin. Both conversions are correct and lead to correct answers if your following steps are also correct. The error is in the step where you simply remove sin from both sides. You will be subject to that error any time you try to "cancel" a sin function on both sides of an equation, regardless of whether there was any "conversion" that got you to that point. – David K Jun 16 '22 at 16:42

3 Answers3

2

If $\sin[\cot^{-1}(x + 1)]=\cos[\tan^{-1}x]$, then

$$\sin[\cot^{-1}(x + 1)]=\sin\left[\frac{\pi}{2}-\tan^{-1}x\right]$$

If $\sin(\alpha) = \sin(\beta)$ then, for some integer, $n$

$$\alpha = \beta + 2n\pi \quad \text{or} \quad \alpha = (\pi - \beta) + 2n\pi$$

So

\begin{align} \cot^{-1}(x + 1) &= \left(\frac{\pi}{2}-\tan^{-1}x\right) + 2n\pi \\ &\text{or} \\ \cot^{-1}(x + 1) &= \left(\frac{\pi}{2}+\tan^{-1}x\right) + 2n\pi \\ \end{align}

You can work out the rest.

Steven Alexis Gregory
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The reason the first method refuses to work is that it gives us $$\arctan x+\text{arccot} (x+1)=\frac π2.$$ But this is wrong, since the arguments of the involved functions are not equal for any value of $x.$

The correct identity is $$\arctan x+\text{arccot}x=\frac π2.$$

Allawonder
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To start with, $$ \cos(\tan^{-1}x) = \frac1{\sqrt{x^2 + 1}} > 0. \tag1 $$

The range of the inverse cotangent is $\left(-\frac\pi2,0\right) \cup \left(0,\frac\pi2\right],$ but from $(1)$ we know that we must have $\sin(\cot^{-1}(x + 1)) > 0$ and this eliminates $-\frac\pi2<\cot^{-1}(x + 1) < 0.$ Therefore $\cot^{-1}(x + 1) > 0$ and $x > -1.$

Under those constraints, $$ \sin(\cot^{-1}(x + 1)) = \frac1{\sqrt{(x + 1)^2 + 1}}. \tag2 $$

Alternatively, if you didn't first observe that $\cot^{-1}(x + 1) > 0$ you could write $(2)$ with a $\pm$ sign in front of the right-hand side, and you should then realize that when equating $(1)$ and $(2)$ there is no possible solution for the $-$ sign so it can be eliminated.

It's then easy to work out the rest of the problem.

This way you avoid all the complications of figuring out which of the infinitely many possible solutions to $\sin(\theta) = y$ are relevant when removing the $\sin$ function from both sides of an equation.

David K
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