Question

Se $tg(x)= \frac{4}{3}$, então: $tg(x/2)= y$

Qual será o valor de y? Explique passo a passo algebricamente.

Answer 1

Temos a seguinte fórmula:

$\boxed{\boxed{\tan(a+b)=\dfrac{\tan a+\tan b}{1-\tan a\cdot\tan b}}}$

Daí, tiramos que

$\tan x=\tan(\frac{x}{2}+\frac{x}{2})=\dfrac{\tan\frac{x}{2}+\tan\frac{x}{2}}{1-\tan\frac{x}{2}\cdot\tan\frac{x}{2}}\\\\\\\tan x=\dfrac{2\tan\frac{x}{2}}{1-\tan^{2}(\frac{x}{2})}\\\\\\\boxed{\boxed{\tan x=\dfrac{2y}{1-y^{2}}}}$

Substituindo $\tan x=\frac{4}{3}$:

$\dfrac{2y}{1-y^{2}}=\dfrac{4}{3}\\\\\\\dfrac{2y}{1-y^{2}}\cdot\dfrac{1}{2}=\dfrac{4}{3}\cdot\dfrac{1}{2}\\\\\\\dfrac{y}{1-y^{2}}=\dfrac{2}{3}$

Assumindo $y^{2}\neq1$ e multiplicando em cruz:

$3y=2(1-y^{2})\\\\3y=2-2y^{2}\\\\2y^{2}+3y-2=0$

Resolveremos a equação por bhaskara e encontraremos os dois valores possíveis para $y$:

$\Delta=b^{2}-4ac\\\Delta=3^{2}-4\cdot2\cdot(-2)\\\Delta=9+16\\\Delta=25\\\\\\y=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-3\pm\sqrt{25}}{2\cdot2}=\dfrac{-3\pm5}{4}=\begin{cases}y_{1}=\dfrac{-3+5}{4}=\dfrac{2}{4}=\dfrac{1}{2}\\\\y_{2}=\dfrac{-3-5}{4}=-\dfrac{8}{4}=-2\end{cases}$

Então, temos dois valores possíveis para $\tan\frac{x}{2}$:

$\boxed{\boxed{\mathsf{\tan x=\frac{4}{3}~~~\Longrightarrow~~\tan\bigg(\frac{x}{2}\bigg)=\frac{1}{2}~~~~~ou~~~~~\tan\bigg(\frac{x}{2}\bigg)=-2}}}$