Question

$\large\tt Calcule:sen\left\{{2~arc~tg\left[\sqrt{3}~cos\left(\dfrac{1}{2}~arc~sen~\dfrac{\sqrt{5}}{3}\right)\right]\right\}$

Answer 1

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$\sf sen\left\{2~arc~tg\left[\sqrt{3}\left(\dfrac{1}{2}~arc~sen~\dfrac{\sqrt{5}}{3}\right)\right]\right\}\\\rm\underline{fac_{\!\!\!,}a}~arc~sen~\dfrac{\sqrt{5}}{3}=\alpha\implies sen~\alpha=\dfrac{\sqrt{5}}{3}\\\sf cos\left(\dfrac{1}{2}~~arc~sen~\dfrac{\sqrt{5}}{3}\right)=cos\left(\dfrac{1}{2}~\alpha\right)=cos\left(\dfrac{\alpha}{2}\right)$

$\sf sen^2~\alpha+cos^2\alpha=1\\\sf \left(\dfrac{\sqrt{5}}{3}\right)^2+cos^2~\alpha=1\\\sf\dfrac{5}{9}+cos^2~\alpha=1\cdot 9\\\sf 5+9cos^2~\alpha=9\\\sf9cos^2~\alpha=9-5\\\sf 9cos^2~\alpha=4\\\sf cos^2~\alpha=\dfrac{4}{9}\\\sf cos~\alpha=\sqrt{\dfrac{4}{9}}\implies cos~\alpha=\dfrac{2}{3}$

$\sf cos\left(\dfrac{\alpha}{2}\right)=\sqrt{\dfrac{1+cos~\alpha}{2}}\\\sf cos\left(\dfrac{\alpha}{2}\right)=\sqrt{\dfrac{1+\frac{2}{3}}{2}}\\\sf cos\left(\dfrac{\alpha}{2}\right)=\sqrt{\dfrac{\frac{5}{3}}{2}}=\sqrt{\dfrac{5}{3}\cdot\dfrac{1}{2}}\\\sf cos\left(\dfrac{\alpha}{2}\right)=\sqrt{\dfrac{5}{6}}$

$\sf sen\left\{2~arc~tg\left[\sqrt{3}\left(\dfrac{1}{2}~arc~sen~\dfrac{\sqrt{5}}{3}\right)\right]\right\}=sen\left\{2~arc~tg\left[\sqrt{3}~\sqrt{\dfrac{5}{6}\right]\right\}\\\sf= sen\left\{2~arc~tg\sqrt{\dfrac{\diagup\!\!\!3\cdot5}{\diagup\!\!\!3\cdot2}}\right\}\\\sf sen\left{2~arc~tg~\sqrt{\dfrac{5}{2}}\right}$

$\sf\underline{\rm fac_{\!\!\!,}a}~arct~tg~\sqrt{\dfrac{5}{2}}=\beta\implies tg~\beta=\sqrt{\dfrac{5}{2}}\\\sf tg^2~\beta+1=\dfrac{1}{cos^2~\beta}\\\sf cos^2~\beta=\dfrac{1}{1+tg^2~\beta}\\\sf tg^2~\beta+1=\left(\sqrt{\dfrac{5}{2}}\right)^2+1\\\sf tg^2~\beta+1=\dfrac{5}{2}+\dfrac{2}{2}=\dfrac{7}{2}\\\sf \dfrac{1}{1+tg^2~\beta}=\dfrac{2}{7}\\\sf cos^2~\beta=\dfrac{2}{7}\\\sf cos~\beta=\sqrt{\dfrac{2}{7}}$

$\sf tg~\beta=\dfrac{sen~\beta}{cos~\beta}\implies sen~\beta=cos~\beta\cdot tg~\beta\\\sf sen~\beta=\sqrt{\dfrac{2}{7}}\cdot\sqrt{\dfrac{5}{2}}\\\sf sen~\beta=\sqrt{\dfrac{\diagup\!\!\!2\cdot5}{7\cdot\diagup\!\!\!2}}\\\sf sen~\beta=\sqrt{\dfrac{5}{7}}$

$\sf sen\left\{2~arc~tg~\sqrt{\dfrac{5}{2}}\right\}=sen~2~\beta\\\sf sen~2\beta=2\cdot sen~\beta\cdot cos~\beta\\\sf sen~2~\beta=2\cdot\sqrt{\dfrac{5}{7}}\cdot\sqrt{\dfrac{2}{7}}\\\sf sen~2\beta=\dfrac{2\sqrt{10}}{7}$

$\boxed{\sf sen\left\{2~arc~tg\left[\sqrt{3}\left(\dfrac{1}{2}~arc~sen~\dfrac{\sqrt{5}}{3}\right)\right]\right\}=\dfrac{2\sqrt{10}}{7}}$