Question

Encontre o valor das expressões:
a) e^2 In 3
b) tg (arcsen 1/2)

Answer 1

a) $e^{2\,\mathrm{\ell n\,}3}$

$=\big(e^{\mathrm{\ell n\,}3}\big)^2~~~~~~~~(\text{mas }e^{\mathrm{\ell n\,}3}=3)\\\\ =3^2\\\\\\ \boxed{\begin{array}{c}e^{2\,\mathrm{\ell n\,}3}=9\end{array}}$

____________

b) $\mathrm{tg}\!\left(\mathrm{arcsen\,}\dfrac{1}{2} \right )$

$\theta=\mathrm{arcsen\,}x~~~~~~\mathbf{(i)}$

com $-1\le x\le 1.$


Portanto,

$x=\mathrm{sen\,}\theta\\\\ x^2=\mathrm{sen^2\,}\theta\\\\ x^2=1-\cos^2 \theta\\\\ \cos^2 \theta=1-x^2\\\\ \cos \theta=\pm \sqrt{1-x^2}~~~~~~\mathbf{(ii)}$


Mas $\theta$ é o $\mathrm{arcsen}$ de algum $x\in [-1,\,1],$ portanto, $\theta$ deve pertencer ao conjunto imagem da função $\mathrm{arcsen}:$

$-\dfrac{\pi}{2}\le \theta\le \dfrac{\pi}{2}$


e neste intervalo, o cosseno de $\theta$ nunca é negativo. Portanto, podemos desconsiderar a raiz quadrada com sinal negativo em $\mathbf{(ii)},$ chegando a

$\cos \theta=\sqrt{1-x^2}\\\\\\ \cos(\mathrm{arcsen\,}x)=\sqrt{1-x^2}~~~~~~\mathbf{(iii)}$

_____

A tangente só vai estar definida se

$-1<x<1$


e neste caso, temos

$\mathrm{tg}(\mathrm{arcsen\,}x)=\dfrac{\mathrm{sen}(\mathrm{arcsen\,}x)}{\cos(\mathrm{arcsen\,}x)}\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathrm{tg}(\mathrm{arcsen\,}x)=\dfrac{x}{\sqrt{1-x^2}} \end{array}}~~~~~~\text{para }-1<x<1$


No caso particular desta questão, temos $x=\dfrac{1}{2}:$

$\mathrm{tg}\!\left(\mathrm{arcsen\,}\dfrac{1}{2}\right)=\dfrac{\frac{1}{2}}{\sqrt{1-\left(\frac{1}{2} \right )^{\!2}}}\\\\\\ =\dfrac{\frac{1}{2}}{\sqrt{1-\frac{1}{4}}}\\\\\\ =\dfrac{\frac{1}{2}}{\sqrt{\frac{4-1}{4}}}\\\\\\ =\dfrac{\frac{1}{2}}{\sqrt{\frac{3}{4}}}\\\\\\ =\dfrac{\left(\frac{1}{2} \right )}{\left(\frac{\sqrt{3}}{2} \right )}\\\\\\ =\dfrac{1}{\diagup\!\!\!\! 2}\cdot \dfrac{\diagup\!\!\!\! 2}{\sqrt{3}}\\\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathrm{tg}\!\left(\mathrm{arcsen\,}\dfrac{1}{2}\right)=\dfrac{1}{\sqrt{3}} \end{array}}$


Bons estudos! :-)