1

$$\arctan a+\arctan\left(\frac 1 a \right)=\frac \pi 2$$

I have the mark scheme in front of me, and I understand where the numbers come from, but I don't understand why they do what they do.

You need this part in to show it: enter image description here

The markscheme says:

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Again, I understand where the numbers come from, but can someone explain to me where the first line comes from, as well as the rest of it? It's a "Hence" question so I have to solve it this way.

Travis Willse
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Jim
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  • Related : http://math.stackexchange.com/questions/610261/simplifying-an-arctan-equation and http://math.stackexchange.com/questions/483465/proof-of-arctan2-pi-2-arctan1-2 – lab bhattacharjee Sep 14 '14 at 16:44
  • Are you looking for a proof using the integral form of arctangent? If not, then the answer below is sufficient. – Semiclassical Sep 14 '14 at 16:46

4 Answers4

4

Notice that the derivative of $\arctan x$ is $\frac1{1+x^2}$ so we have

$$\int_1^\alpha \frac{dx}{1+x^2}=\arctan x\Bigg|_1^\alpha$$ now use the change of variable $t=\frac1x$ to find the given equality. Can you take it from here?

2

This is too long for a comment, but here's a method that uses only geometry and the definition of $\arctan$: For $a > 0$, draw a right triangle with legs $1$ and $a$ adjacent to the right angle. Then, the two acute angles are, by definition $\arctan a$ and $\arctan \left( \frac{1}{a} \right)$, and since the triangle is right, the sum of these two angles is $\frac{\pi}{2}$.

Remark Note that your identity holds only for positive $a$. For negative $a$ the sum is equal to $-\frac{\pi}{2}$, which you can show by applying the identity to $|a|$ and using that $\arctan$ is odd.

Travis Willse
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1

Differentiate with respect to $a$, to prove that it has to be constant. Then taking the value at $a=1$ gives you the result. That's probably the easiest way to prove it.

Matt B.
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  • Strictly speaking, this only shows that the quantity is constant on the connected component $(0, \infty)$ of the domain, and in fact it does not hold on the other component. – Travis Willse Sep 14 '14 at 16:41
  • If $a>0$ then $\frac{1}{a}>0$ so indeed it works on $(0,+\infty)$, same calculation for $a<0$ and you can take it on $a=-1$. – Matt B. Sep 14 '14 at 16:44
1

Supose $a\neq 0$. It's just a way to remember:$$\tan\left(\arctan a+\arctan\frac{1}{a}\right)=\frac{\tan(\arctan a)+\tan\left(\arctan\frac{1}{a}\right)}{1-\tan(\arctan a)\arctan\left(\frac{1}{a}\right)}=\frac{a+\frac{1}{a}}{1-a\cdot \frac{1}{a}}=+\infty. $$ Moreover, $\arctan(+\infty )=\frac{\pi}{2}$

This proof is not correct (but funny anyway :-) )

idm
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