Question

calcular tg ( 2 arctg 1/5)

Answer 1

Seja $\theta=\mathrm{arctg\,}\frac{1}{5}.$ Então,

$\mathrm{tg\,}\theta=\mathrm{tg}\left(\mathrm{arctg}\frac{1}{5} \right )\\ \\ \mathrm{tg\,}\theta=\frac{1}{5}$


Sendo assim, queremos calcular,

$\mathrm{tg\,}(2\theta)$


Pela identidade da tangente do arco duplo, temos que

$\mathrm{tg\,}(2\theta)=\dfrac{2\,\mathrm{tg\,}\theta}{1-\mathrm{tg^{2}\,}\theta}\\ \\ \\ \mathrm{tg\,}(2\theta)=\dfrac{2\cdot \frac{1}{5}}{1-\left(\frac{1}{5} \right )^{2}}\\ \\ \\ \mathrm{tg\,}(2\theta)=\dfrac{\frac{2}{5}}{1-\frac{1}{25}}\\ \\ \\ \mathrm{tg\,}(2\theta)=\dfrac{\frac{2}{5}}{\left(\frac{25-1}{25} \right )}\\ \\ \\ \mathrm{tg\,}(2\theta)=\dfrac{\left(\frac{2}{5} \right )}{\left(\frac{24}{25} \right )}\\ \\ \\ \mathrm{tg\,}(2\theta)=\frac{2}{5}\cdot \frac{25}{24}\\ \\ \mathrm{tg\,}(2\theta)=\frac{5}{12}\\ \\ \\ \Rightarrow\;\;\boxed{\begin{array}{c} \mathrm{tg\,}(2\,\mathrm{arctg\,}\frac{1}{5})=\frac{5}{12} \end{array}}$