Trigonometry Identity Domain

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).

Given that the arctangent function is restricted to the principle branch of (π2,π2) (- \frac{\pi}{2} , \frac{\pi}{2} ) . What is the actual domain for the arctangent identity? Can you find sufficient and necessary conditions? How do we show that these conditions are indeed sufficient and necessary?


This question arose from observing comments in Trigonometry Paradox, and seeing that various people had different sets which they felt would work. For example:

Derek - xy<1 xy < 1 .
Heli - x>0,y>0,xy<1 x> 0, y > 0 , xy<1 .
Aditya - Sum of LHS lies in range.

Note by Calvin Lin
6 years, 1 month ago

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For what it's worth, Derek's condition xy<1 xy < 1 and Aditya's condition π/2<tan1x+tan1y<π/2 -\pi/2 < \tan^{-1} x + \tan^{-1} y < \pi/2 (where the range of the inverse tangent is taken to be the open interval (π/2,π/2) (-\pi/2, \pi/2) ) are equivalent, necessary, and sufficient. Heli's condition (assuming all components must hold) clearly is not equivalent to the others, being a proper subset, and the counterexample x=y=1/3 x = y = -1/\sqrt{3} shows it is not necessary.

I will leave the proof for other students to explore. It's a good exercise.

hero p. - 6 years, 1 month ago

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Em I think how I understand this is like this If tan(x)=y then arctan(y)=d With d=x+2k*pi ( k is an integer ) and d is in (-pi/2;pi/2) We can apply the arctan t about any real number.

Ahmed Taha - 6 years, 1 month ago

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The arctangent function is defined to give a single value (which lies in the principle branch).

The 'alternative interpretation' of arctangent is that of a multivalued function, in which case the equality condition would be interpreted as a particular solution of the multivalued function (i.e. with a correct choice of variables kik_i). However, students often do not have a good understanding of this concept, nor do they know what correct choice of variables work, which is why the above conundrum arises. Alternatively, you could interpret this question as "For what values of xx and yy, does the equation hold with kx=ky=0k_x = k_y = 0 ?"

Note: I believe you want d=x+kπ d = x + k\pi instead of d=x+2kπd = x + 2k\pi, since we're working with tangent which has a period of π\pi.

Calvin Lin Staff - 6 years, 1 month ago

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Ahh yes , I often get confused. If ine just memorises the graph of the Arctan , it will help alot , and the variable work will only be an application f one's degree of understanding.

Ahmed Taha - 6 years, 1 month ago

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