¿Cómo puede $$\pi - \arctan(\frac{8}{x}) - \arctan(\frac{13}{20-x}) = \arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})$$
Respuestas
¿Demasiados anuncios?Utilice el hecho de que para $t\not=0$ , $\displaystyle \arctan(t)+\arctan(1/t)=\mbox{sgn}(t)\cdot\frac{\pi}{2}$ .
Consulte aquí los detalles: $\arctan (x) + \arctan(1/x) = \frac{\pi}{2}$
Entonces para $x\not=0$ y $x\not=20$ , $$\arctan(\frac{8}{x})+\arctan(\frac{13}{20-x})+\arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})\\ =\mbox{sgn}(x)\cdot\frac{\pi}{2}+ \mbox{sgn}(20-x)\cdot\frac{\pi}{2}=\begin{cases} \pi\quad \mbox{if $0<x<20$,}\\ 0 \quad \mbox{if $x<0$ or $x>20$.}\\ \end{cases}$$
Existe una identidad trigonométrica para $\arctan$ : $\arctan(\alpha) \pm \arctan(\beta) = \arctan(\frac{\alpha \pm \beta}{1 \mp \alpha\beta})$ .
En nuestro caso:
$$ \pi-\arctan(\frac{8}{x})-\arctan(\frac{13}{20-x})=\arctan(\frac{x}{8})+\arctan(\frac{20-x}{13})\\ \pi-\arctan(\frac{13}{20-x})-\arctan(\frac{20-x}{13})=\arctan(\frac{8}{x})+\arctan(\frac{x}{8})\\ \pi-(\arctan(\frac{13}{20-x})+\arctan(\frac{20-x}{13}))=\arctan(\frac{8}{x})+\arctan(\frac{x}{8})\\ \pi-\arctan(\frac{\frac{13}{20-x}+\frac{20-x}{13}}{1-(\frac{13}{20-x}\cdot \frac{20-x}{13})})=\arctan(\frac{\frac{8}{x}+\frac{x}{8}}{1-(\frac{8}{x}\cdot \frac{x}{8})})\\ \pi-\arctan(\frac{13^2+(20-x)^2}{0})=\arctan(\frac{8^2+x^2}{0})\\ $$ Obsérvese que el ángulo para $lim_{x\rightarrow \infty} \arctan(x) =\frac{\pi}{2}$ (en caso de que $x<0$ obtenemos $-\frac{\pi}{2})$ .
por lo tanto, obtenemos: $\pi - \frac{\pi}{2}=\frac{\pi}{2}$ .
$$\frac{\pi}{2} - \arctan\left(\frac{8}{x}\right) + \frac{\pi}{2} - \arctan\left(\frac{13}{20-x}\right) = \arctan\left(\frac{x}{8}\right) + \arctan\left(\frac{20-x}{13}\right)$$ $$\mbox{arccot} \left(\frac{8}{x}\right) +\mbox{arccot}\left(\frac{13}{20-x}\right)=\arctan\left(\frac{x}{8}\right) + \arctan\left(\frac{20-x}{13}\right)$$
$$\color{red}{\boxed{ \color{black}{\mbox{arccot} \left(\frac{1}{t}\right) =\arctan t}}} $$ $\color{red}{\text{This is valid only when}\qquad t \gt 0}$