The sum of tangents trigonometric identity gives us that

$\tan ( \alpha + \beta) = \frac{ \tan \alpha + \tan \beta} { 1 - \tan \alpha \tan \beta} .$

By letting $\alpha = \tan^{-1} x$ and $\beta = \tan^{-1} y$, the equivalent trigonometric identity on $\tan^{-1}$ is

$\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y} { 1-xy} \right).$

Let's use this identity to calculate $\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3$. We first have

$\tan^{-1} 1 + \tan^{-1} 2 = \tan^{-1} \left( \frac { 1 + 2 } { 1 - 1 \times 2} \right)= \tan^{-1} (-3).$

As such, this gives

\begin{aligned} & \tan^{-1} 1+ \tan^{-1} 2 + \tan^{-1} 3 \\ = & \tan^{-1} (-3) + \tan^{-1} 3\\ = & \tan^{-1} \left( \frac{-3 + 3} { 1 - (-3)\times 3 } \right) \\ = & \tan^{-1} 0 \\ = & 0 \\ \end{aligned}

By considering the corresponding right triangles, we get that $\tan^{-1} 1 > 0$, $\tan^{-1} 2 > 0$ and $\tan^{-1} 3 > 0$. Hence, the sum of 3 positive terms is 0.

What went wrong?

This arose when I was reviewing the solution of a submitted problem. Note by Calvin Lin
6 years, 4 months ago

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Note that $\alpha, \beta$ is in the range $(-\frac{\pi}{2},-\frac{\pi}{2})$ and that $\tan \alpha = x$ and $\tan \beta = y$. So $\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \Rightarrow \tan (\tan^{-1}x + \tan^{-1}y) = \frac{x + y}{1 - xy}$. We cannot just simply take $\tan^{-1}$ on both sides: This operation is only valid when $\tan^{-1}x + \tan^{-1}y = \alpha + \beta$ is in the range $(-\frac{\pi}{2},-\frac{\pi}{2})$, which might not necessarily be the case.

But of course, if we restrict the range of $x,y$ to be in $(-1,1)$, then $\tan^{-1}x, \tan^{-1}y$ will be in the range $(-\frac{\pi}{4},-\frac{\pi}{4})$, so their sum would be in the range $(-\frac{\pi}{2},-\frac{\pi}{2})$ and the equation would hold. More generally, we could also say that $\tan^{-1}x + \tan^{-1}y = \tan^{-1} (\frac{x + y}{1 - xy}) \pmod{\pi}$ (where $xy \neq 1$). Note that indeed, we have $\pi = \tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3 \equiv 0 \pmod{\pi}$.

- 6 years, 4 months ago

It is in fact possible to prove that $\alpha + \beta$ is in the range $(-\frac{\pi}{2},\frac{\pi}{2})$ iff $xy < 1$, so $\tan^{-1} x + \tan^{-1} y = \tan^{-1} (\frac{x+y}{1-xy})$ if $xy < 1$. It is also not difficult to show that $\tan^{-1} x + \tan^{-1} y = \tan^{-1} (\frac{x+y}{1-xy}) + \pi$ if $xy > 1$ and both $x, y$ are positive, while $\tan^{-1} x + \tan^{-1} y = \tan^{-1} (\frac{x+y}{1-xy}) - \pi$ if $xy > 1$ and both $x,y$ are negative.

- 6 years, 4 months ago

The key reason for this behavior is the fact that the trigonometric functions are periodic, and therefore, not injective. Hence their inverse mappings are not functions unless they are restricted to a particular branch.

A simple analogy is the behavior of the function $f : \mathbb{R} \to \mathbb{R}, f(x) = x^2$. The inverse mapping, $f^{-1}(x) = x^{1/2}$, has two images for each nonzero $x$, because $f(-x) = f(x)$ for all $x$. Similarly, because $\tan(\theta + k \pi) = \tan \theta$ for all integers $k$, there are infinitely many angles whose tangent is equal to some given value.

It is possible to regard the inverse tangent as a mapping of a number to a set: for instance, if $\tan \theta = z$, then $\tan^{-1} z = \{ \theta + k \pi : k \in \mathbb{Z} \}$. Then there is no contradiction, because $\tan^{-1} 0 = \{ \ldots, -2\pi, -\pi, 0, \pi, 2\pi, \ldots \}$.

Further consideration of the above leads us to conclude that the claim that $\tan^{-1} 1 > 0$ contains an unstated assumption, namely that a particular branch of the inverse tangent is chosen (e.g., $-\pi/2 < \tan^{-1} z < \pi/2$ ). But of course, $\tan -3\pi /4 = 1$ just as much as $\tan \pi/4 = 1$.

From a group- or number-theoretic standpoint, we are actually looking at equivalence classes of the inverse tangent, modulo $\pi$. If $a + b \equiv c \pmod m$, that does not necessarily mean that $a + b = c$, because $a, b, c$ are representatives of their respective equivalence classes modulo $m$.

- 6 years, 4 months ago

Trigometry? I suppose Trigonometry?

- 6 years, 4 months ago

Stop highlighting grammatical errors stupidly.

- 6 years, 4 months ago

You can't use this formula for arctanx and arctany if xy>1, which in this case, holds as 2x1=2>1. You must write pi + arctan1+arctan2+arctan3 = pi.

- 6 years, 4 months ago

Why must we write it that way? How do you know that it is not equal to $-\pi$ or $2 \pi$ or something else?

Staff - 6 years, 4 months ago

Its well known that atan(1)+atan(2)+atan(3)=pi. The issue occurs because of the restricted range of the arc tangent function.

- 6 years, 4 months ago

because in inverse trigometry function range of tan−1 is from ]−π2,π2[ i.e, tan−12&tan−13 not follow the rule of tan−1x+tan−1y=tan−1(x+y/1−xy).

- 6 years, 4 months ago

Why doesn't it follow the rule? Those values still lie in the range of the inverse trigonometric function.

Staff - 6 years, 4 months ago

$\tan^-1 x + \tan^-1 y = \tan^-1 (\frac{x + y}{1 - xy})$ if $xy < 1$, which doesn't hold in the first case itself,when you evaluated $\tan^-1 (-3)$. Again, $\tan^-1 x + \tan^-1 y = \pi + \tan^-1 (\frac{x + y}{1 - xy})$ if $xy > 1$, which is the appropriate correction to this paradox. These relations can be easily verified, by considering the quadrants & simple manipulation.

- 6 years, 4 months ago

You're right about the part where $\tan^{-1} x + \tan^{-1} y = \tan^{-1} (\frac{x+y}{1-xy})$ if $xy < 1$. But one thing to note: $\tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1} (\frac{x+y}{1-xy})$ holds only if $xy > 1$ and both $x,y$ are positive. It does not hold when both $x,y$ are negative. When both $x,y$ are negative and we have $xy > 1$, we should use $\tan^{-1} x + \tan^{-1} y = \tan^{-1} (\frac{x+y}{1-xy}) - \pi$ instead.

- 6 years, 4 months ago

Simply speaking, the "equivalent trigonometric identity" mentioned is not in fact, an identity at all. This is so because the range of the left side is greater and may exceed the defined range of arctan function on the right. Therefore, it is an identity only while the sum on the left side lies in the range defined for the arctan function i.e. (−π/2,π/2).

- 6 years, 4 months ago

How certain are you that your characterization of "it is an identity only while ..." is true? Are there potentially cases that you missed out? Note that many people are looking at the cases where $xy<1$. Do they know something that you don't?

Staff - 6 years, 4 months ago

We use the fact that $\tan^{-1}x=\frac{\pi}{2}-\tan^{-1}(\frac{1}{x})$. $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3 =\frac{\pi}{4}+\frac{\pi}{2}-\tan^{-1}(\frac{1}{2})+\frac{\pi}{2}-\tan^{-1}(\frac{1}{3})$. As $\displaystyle \frac{1}{2} \times \frac{1}{3} < 1, \tan^{-1}\left(\frac{1}{2}\right)+tan^{-1}\left(\frac{1}{3}\right)=\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\times\frac{1}{3}}\right)=\tan^{-1}1=\frac{\pi}{4}$ Therefore, $\displaystyle\tan^{-1}1+\tan^{-1}2+\tan^{-1}3 =\frac{\pi}{4}+\frac{\pi}{2}-\tan^{-1}\left(\frac{1}{2}\right)+\frac{\pi}{2}-\tan^{-1}\left(\frac{1}{3}\right)$ $\displaystyle= \pi + \frac{\pi}{4} - \left(\tan^{-1}\left(\frac{1}{2}\right)+tan^{-1}\left(\frac{1}{3}\right)\right)=\pi$.

- 6 years, 4 months ago

But the range for the function $tan^{-1} x$ is $(-\frac {\pi}{2}, \frac {pi}{2})$.$(\frac {\pi}{2}) \leq 2$, so we must subtract integral multiples of $\pi$ to make it in the range $-(\frac {\pi}{2}, \frac {\pi}{2})$.

Thus, $tan ^{-1} 1 + tan^{-1} 2 + tan^{-1} 3$ = $tan^{-1} 1 + tan ^{-1} (2-\pi) + tan^{-1} (3-\pi)$ = $\pi$.

- 6 years, 4 months ago

You have things mixed up. $\tan ^{-1} 2 \neq \tan^{-1} (2- \pi)$. You are thinking of $\tan \theta$ instead, which is periodic with period $\pi$.

Staff - 6 years, 4 months ago

this statement tan−1x+tan−1y=tan−1(x+y1−xy) is true,. iff x>0,y>0 and most imp xy<1...........

- 6 years, 4 months ago

How certain are you about the if and only if claim? Does it hold for $x = - r, y = r$, where $r$ is any real number?

Note that the answer will (may) change according to what the restricted domain is. For sake of clarity, let's stick to $[ - \frac{\pi}{2} , \frac{ \pi } { 2} )$.

Staff - 6 years, 4 months ago

Ah. Was this part of the submission of the Russell's triple tangent problem this week?

- 6 years, 4 months ago

No, the original problem was rejected, because this represented a huge hole in the argument.

Staff - 6 years, 4 months ago

geometrically solving this could probably avoid confusion.

- 6 years, 4 months ago

yeah..perhaps..=((

- 6 years, 4 months ago

Please answer my Q G is a group a is an element of order 5 and x is an element of order 2 what is order of x inverse

- 6 years, 4 months ago

Well, $\tan^{-1} 0$ does not necessarily equal $0$. It depends on the interval of the angles you want the solution to be in.

- 6 years, 4 months ago