Question

Calcule o valor de Sec a/2, dado Sen a=1/3, com arco a pertencente ao 1 quadrante

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Answer 1

$\large\boxed{\begin{array}{l}\sf sen(a)=\dfrac{1}{3}\implies sen^2(a)=\dfrac{1}{9}\\\\\sf cos^2(a)=1-sen^2(a)\\\sf cos^2(a)=\dfrac{9}{9}-\dfrac{1}{9}=\dfrac{8}{9}\\\\\sf cos(a)=\dfrac{\sqrt{8}}{\sqrt{9}}=\dfrac{2\sqrt{2}}{3}\\\sf cos\bigg(\dfrac{a}{2}\bigg)=\sqrt{1+cos(a)}\\\sf cos\bigg(\dfrac{a}{2}\bigg)=\sqrt{1+\dfrac{2\sqrt{2}}{3}}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Radical\,duplo}\\\sf\sqrt{A\pm\sqrt{B}}=\sqrt{\dfrac{A+C}{2}}\pm\sqrt{\dfrac{A-C}{2}}\\\sf onde~C=\sqrt{A^2-B}\\\sf\dfrac{2\sqrt{2}}{3}=\sqrt{\bigg(\dfrac{2}{3}\bigg)^2\cdot2}=\sqrt{\dfrac{8}{9}}\\\sf\sqrt{1+\sqrt{\dfrac{8}{9}}}\\\\\sf C=\sqrt{1^2-\dfrac{8}{9}}=\sqrt{\dfrac{9}{9}-\dfrac{8}{9}}=\dfrac{1}{3}\end{array}}$

$\large\boxed{\begin{array}{l}\sf\sqrt{1+\sqrt{\dfrac{8}{9}}}=\sqrt{\dfrac{1+\frac{1}{3}}{2}}+\sqrt{\dfrac{1-\frac{1}{3}}{2}}\\\\\sf\sqrt{1+\sqrt{\dfrac{8}{9}}}=\sqrt{\dfrac{1}{\diagup\!\!\!2}\cdot\dfrac{\diagup\!\!\!4^2}{3}}+\sqrt{\dfrac{1}{\diagup\!\!\!2}\cdot\dfrac{\diagup\!\!\!2}{3}}\\\\\sf\sqrt{1+\sqrt{\dfrac{8}{9}}}=\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{6}}{3}+\dfrac{\sqrt{3}}{3}\\\\\sf\sqrt{1+\sqrt{\dfrac{8}{9}}}=\dfrac{\sqrt{6}+\sqrt{3}}{3}\end{array}}$

$\large\boxed{\begin{array}{l}\sf\sqrt{1+\dfrac{2\sqrt{2}}{3}}=\sqrt{1+\sqrt{\dfrac{8}{9}}}=\dfrac{\sqrt{6}+\sqrt{3}}{3}\\\\\sf cos\bigg(\dfrac{a}{2}\bigg)=\dfrac{\sqrt{6}+\sqrt{3}}{3}\\\\\sf sec\bigg(\dfrac{a}{2}\bigg)=\dfrac{1}{cos\bigg(\dfrac{a}{2}\bigg)}\\\\\sf sec\bigg(\dfrac{a}{2}\bigg)=\dfrac{3}{\sqrt{6}+\sqrt{3}}\\\\\sf sec\bigg(\dfrac{a}{2}\bigg)=\dfrac{3\cdot(\sqrt{6}-\sqrt{3})}{(\sqrt{6}+\sqrt{3})(\sqrt{6}-\sqrt{3})}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf sec\bigg(\dfrac{a}{2}\bigg)=\dfrac{3(\sqrt{6}-\sqrt{3})}{\sqrt{6^2}-\sqrt{3^2}}\\\\\sf sec\bigg(\dfrac{a}{2}\bigg)=\dfrac{3(\sqrt{6}-\sqrt{3})}{6-3}\\\\\sf sec\bigg(\dfrac{a}{2}\bigg)=\dfrac{\backslash\!\!\!3(\sqrt{6}-\sqrt{3})}{\backslash\!\!\!3}\\\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\sf sec\bigg(\dfrac{a}{2}\bigg)=\sqrt{6}-\sqrt{3}}}}}}\end{array}}$