Trigonometry Paradox

The sum of tangents trigonometric identity gives us that

tan(α+β)=tanα+tanβ1tanαtanβ. \tan ( \alpha + \beta) = \frac{ \tan \alpha + \tan \beta} { 1 - \tan \alpha \tan \beta} .

By letting α=tan1x \alpha = \tan^{-1} x and β=tan1y \beta = \tan^{-1} y , the equivalent trigonometric identity on tan1 \tan^{-1} is

tan1x+tan1y=tan1(x+y1xy). \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y} { 1-xy} \right).

Let's use this identity to calculate tan11+tan12+tan13 \tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3 . We first have

tan11+tan12=tan1(1+211×2)=tan1(3). \tan^{-1} 1 + \tan^{-1} 2 = \tan^{-1} \left( \frac { 1 + 2 } { 1 - 1 \times 2} \right)= \tan^{-1} (-3).

As such, this gives

tan11+tan12+tan13=tan1(3)+tan13=tan1(3+31(3)×3)=tan10=0 \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 tan11>0 \tan^{-1} 1 > 0 , tan12>0 \tan^{-1} 2 > 0 and tan13>0 \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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22 votes

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Note that α,β\alpha, \beta is in the range (π2,π2)(-\frac{\pi}{2},-\frac{\pi}{2}) and that tanα=x\tan \alpha = x and tanβ=y\tan \beta = y. So tan(α+β)=tanα+tanβ1tanαtanβtan(tan1x+tan1y)=x+y1xy\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 tan1\tan^{-1} on both sides: This operation is only valid when tan1x+tan1y=α+β\tan^{-1}x + \tan^{-1}y = \alpha + \beta is in the range (π2,π2)(-\frac{\pi}{2},-\frac{\pi}{2}), which might not necessarily be the case.

But of course, if we restrict the range of x,yx,y to be in (1,1)(-1,1), then tan1x,tan1y\tan^{-1}x, \tan^{-1}y will be in the range (π4,π4)(-\frac{\pi}{4},-\frac{\pi}{4}), so their sum would be in the range (π2,π2)(-\frac{\pi}{2},-\frac{\pi}{2}) and the equation would hold. More generally, we could also say that tan1x+tan1y=tan1(x+y1xy)(modπ)\tan^{-1}x + \tan^{-1}y = \tan^{-1} (\frac{x + y}{1 - xy}) \pmod{\pi} (where xy1xy \neq 1). Note that indeed, we have π=tan11+tan12+tan130(modπ)\pi = \tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3 \equiv 0 \pmod{\pi}.

Derek Khu - 6 years, 4 months ago

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It is in fact possible to prove that α+β\alpha + \beta is in the range (π2,π2)(-\frac{\pi}{2},\frac{\pi}{2}) iff xy<1xy < 1, so tan1x+tan1y=tan1(x+y1xy)\tan^{-1} x + \tan^{-1} y = \tan^{-1} (\frac{x+y}{1-xy}) if xy<1xy < 1. It is also not difficult to show that tan1x+tan1y=tan1(x+y1xy)+π\tan^{-1} x + \tan^{-1} y = \tan^{-1} (\frac{x+y}{1-xy}) + \pi if xy>1xy > 1 and both x,yx, y are positive, while tan1x+tan1y=tan1(x+y1xy)π\tan^{-1} x + \tan^{-1} y = \tan^{-1} (\frac{x+y}{1-xy}) - \pi if xy>1xy > 1 and both x,yx,y are negative.

Derek Khu - 6 years, 4 months ago

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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:RR,f(x)=x2 f : \mathbb{R} \to \mathbb{R}, f(x) = x^2 . The inverse mapping, f1(x)=x1/2 f^{-1}(x) = x^{1/2} , has two images for each nonzero x x , because f(x)=f(x) f(-x) = f(x) for all x x . Similarly, because tan(θ+kπ)=tanθ \tan(\theta + k \pi) = \tan \theta for all integers k 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 tanθ=z \tan \theta = z , then tan1z={θ+kπ:kZ} \tan^{-1} z = \{ \theta + k \pi : k \in \mathbb{Z} \} . Then there is no contradiction, because tan10={,2π,π,0,π,2π,} \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 tan11>0 \tan^{-1} 1 > 0 contains an unstated assumption, namely that a particular branch of the inverse tangent is chosen (e.g., π/2<tan1z<π/2 -\pi/2 < \tan^{-1} z < \pi/2 ). But of course, tan3π/4=1 \tan -3\pi /4 = 1 just as much as tanπ/4=1 \tan \pi/4 = 1 .

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

hero p. - 6 years, 4 months ago

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Trigometry? I suppose Trigonometry?

Tim Vermeulen - 6 years, 4 months ago

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Stop highlighting grammatical errors stupidly.

Shourya Pandey - 6 years, 4 months ago

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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.

Leonardo DiCaprio - 6 years, 4 months ago

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

Calvin Lin Staff - 6 years, 4 months ago

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

Samir Khan - 6 years, 4 months ago

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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).

shreyansh gupta - 6 years, 4 months ago

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

Calvin Lin Staff - 6 years, 4 months ago

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tan1x+tan1y=tan1(x+y1xy)\tan^-1 x + \tan^-1 y = \tan^-1 (\frac{x + y}{1 - xy}) if xy<1xy < 1, which doesn't hold in the first case itself,when you evaluated tan1(3)\tan^-1 (-3). Again, tan1x+tan1y=π+tan1(x+y1xy)\tan^-1 x + \tan^-1 y = \pi + \tan^-1 (\frac{x + y}{1 - xy}) if xy>1xy > 1, which is the appropriate correction to this paradox. These relations can be easily verified, by considering the quadrants & simple manipulation.

A Brilliant Member - 6 years, 4 months ago

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You're right about the part where tan1x+tan1y=tan1(x+y1xy)\tan^{-1} x + \tan^{-1} y = \tan^{-1} (\frac{x+y}{1-xy}) if xy<1xy < 1. But one thing to note: tan1x+tan1y=π+tan1(x+y1xy)\tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1} (\frac{x+y}{1-xy}) holds only if xy>1xy > 1 and both x,yx,y are positive. It does not hold when both x,yx,y are negative. When both x,yx,y are negative and we have xy>1xy > 1, we should use tan1x+tan1y=tan1(x+y1xy)π\tan^{-1} x + \tan^{-1} y = \tan^{-1} (\frac{x+y}{1-xy}) - \pi instead.

Derek Khu - 6 years, 4 months ago

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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).

Aditya Kumar - 6 years, 4 months ago

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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 xy<1. Do they know something that you don't?

Calvin Lin Staff - 6 years, 4 months ago

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We use the fact that tan1x=π2tan1(1x)\tan^{-1}x=\frac{\pi}{2}-\tan^{-1}(\frac{1}{x}). tan11+tan12+tan13=π4+π2tan1(12)+π2tan1(13)\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 12×13<1,tan1(12)+tan1(13)=tan1(12+13112×13)=tan11=π4\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, tan11+tan12+tan13=π4+π2tan1(12)+π2tan1(13)\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) =π+π4(tan1(12)+tan1(13))=π\displaystyle= \pi + \frac{\pi}{4} - \left(\tan^{-1}\left(\frac{1}{2}\right)+tan^{-1}\left(\frac{1}{3}\right)\right)=\pi.

Eklavya Sharma - 6 years, 4 months ago

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But the range for the function tan1xtan^{-1} x is (π2,pi2)(-\frac {\pi}{2}, \frac {pi}{2}).(π2)2(\frac {\pi}{2}) \leq 2, so we must subtract integral multiples of π\pi to make it in the range (π2,π2)-(\frac {\pi}{2}, \frac {\pi}{2}).

Thus, tan11+tan12+tan13tan ^{-1} 1 + tan^{-1} 2 + tan^{-1} 3 = tan11+tan1(2π)+tan1(3π)tan^{-1} 1 + tan ^{-1} (2-\pi) + tan^{-1} (3-\pi) = π\pi.

Shourya Pandey - 6 years, 4 months ago

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You have things mixed up. tan12tan1(2π) \tan ^{-1} 2 \neq \tan^{-1} (2- \pi) . You are thinking of tanθ \tan \theta instead, which is periodic with period π \pi .

Calvin Lin Staff - 6 years, 4 months ago

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

heli trivedi - 6 years, 4 months ago

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How certain are you about the if and only if claim? Does it hold for x=r,y=r x = - r, y = r , where rr 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 [π2,π2) [ - \frac{\pi}{2} , \frac{ \pi } { 2} ) .

Calvin Lin Staff - 6 years, 4 months ago

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

Tanishq Aggarwal - 6 years, 4 months ago

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

Calvin Lin Staff - 6 years, 4 months ago

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geometrically solving this could probably avoid confusion.

Daniel Wang - 6 years, 4 months ago

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yeah..perhaps..=((

Bonaventura Radityo Sanjoyo - 6 years, 4 months ago

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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

sai venkata raju nanduri - 6 years, 4 months ago

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Well, tan10\tan^{-1} 0 does not necessarily equal 0 0 . It depends on the interval of the angles you want the solution to be in.

Dimitrij Ray Susantio - 6 years, 4 months ago

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