Calculate $ \arctan(x_{1}) \cdot \arctan(x_{2}) $

Question

Let $ x_{1} $ and $ x_{2} $ be the roots of the equation : $ x^2-2\sqrt{2}x+1=0 $

Calculate $ \arctan(x_{1}) \cdot \arctan(x_{2}) $

The answer should be $ \dfrac{3\pi^2}{64} $.

How does the fact that $ x_{1} $ = $ 1 + \sqrt2 $ = $ \dfrac{1}{x_{2}} $ help?

Answer 1

We have $$x^2-2\sqrt{2}x+1=(x-(1+\sqrt2))(x-(-1+\sqrt2))=0$$ so let $x_1=1+\sqrt2$ and $x_2=-1+\sqrt2$.

Let $a=\tan^{-1}(1+\sqrt2)$. Using the double angle tangent formula, $$\begin{align}\tan2a=\frac{2\tan a}{1-\tan^2a}&\implies\tan2a=\frac{2(1+\sqrt2)}{1-(1+\sqrt2)^2}=-1\end{align}$$ and hence taking the principal angle gives $$\tan^{-1}(1+\sqrt2)=\frac38\pi$$

Let $b=\tan^{-1}(-1+\sqrt2)$. Using the double angle tangent formula, $$\begin{align}\tan2b=\frac{2\tan b}{1-\tan^2b}&\implies\tan2b=\frac{2(-1+\sqrt2)}{1-(-1+\sqrt2)^2}=1\end{align}$$ and hence taking the principal angle gives $$\tan^{-1}(-1+\sqrt2)=\frac18\pi$$ Therefore $$\boxed{\tan^{-1}(x_1)\cdot\tan^{-1}(x_2)=\frac38\pi\cdot\frac18\pi=\frac3{64}\pi^2}$$ as desired.

Answer 2

$$\csc2t-\cot2t=\tan t$$

Here $2t=2m\pi + \dfrac\pi4$ where $m$ is any integer as $\csc2t =\sqrt2=\csc\dfrac\pi4$ and $\cot2t=1=\cot?$

But $-\dfrac\pi2<\arctan(\tan t)<\dfrac\pi2$ using https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values

Similarly, $$\csc2t+\cot2t=\cot t=\tan\left(\dfrac\pi2-t\right)$$

Answer 3

use that $$\arctan(\sqrt{2}+1)=\frac{3}{8}\pi$$