Question

Calcule: sen r/8, cos r/8 e tg r/8

Answer 1

Utilizaremos as fórmulas do arco-metade:

$\mathrm{sen^{2}}\left(\dfrac{\theta}{2} \right )=\dfrac{1-\cos \theta}{2}\\ \\ \cos^{2} \left(\dfrac{\theta}{2} \right )=\dfrac{1+\cos \theta}{2}$


Se fizermos $\theta=\dfrac{\pi}{4}$, temos que

$\dfrac{\theta}{2}=\dfrac{^{\pi}\!\!\!\diagup\!\!_{4}}{2}\\ \\ \dfrac{\theta}{2}=\dfrac{\pi}{4}\cdot \dfrac{1}{2}\\ \\ \dfrac{\theta}{2}=\dfrac{\pi}{8}$


Substituiremos nas fórmulas os valores dos ângulos.


Encontrando o seno,

$\bullet\;\;\mathrm{sen^{2}}\left(\dfrac{\pi}{8} \right )=\dfrac{1-\cos \frac{\pi}{4}}{2}\\ \\ \mathrm{sen^{2}}\left(\dfrac{\pi}{8} \right )=\dfrac{1-\frac{\sqrt{2}}{2}}{2}\\ \\ \mathrm{sen^{2}}\left(\dfrac{\pi}{8} \right )=\dfrac{2-\sqrt{2}}{2}\cdot \dfrac{1}{2}\\ \\ \mathrm{sen^{2}}\left(\dfrac{\pi}{8} \right )=\dfrac{2-\sqrt{2}}{4}\\ \\ \mathrm{sen}\left(\dfrac{\pi}{8} \right )=\pm \sqrt{\dfrac{2-\sqrt{2}}{4}}\\ \\ \mathrm{sen}\left(\dfrac{\pi}{8} \right )=\pm \dfrac{\sqrt{2-\sqrt{2}}}{2}$



Como $\dfrac{\pi}{8}$ é um arco do primeiro quadrante, o seu seno é positivo. Logo,

$\mathrm{sen}\left(\dfrac{\pi}{8} \right )=\dfrac{\sqrt{2-\sqrt{2}}}{2}$


Encontrando o cosseno,

$\bullet\;\;\cos^{2} \left(\dfrac{\pi}{8} \right )=\dfrac{1+\cos \frac{\pi}{4}}{2}\\ \\ \cos^{2}\left(\dfrac{\pi}{8} \right )=\dfrac{1+\frac{\sqrt{2}}{2}}{2}\\ \\ \cos^{2}\left(\dfrac{\pi}{8} \right )=\dfrac{2+\sqrt{2}}{2}\cdot \dfrac{1}{2}\\ \\ \cos^{2}\left(\dfrac{\pi}{8} \right )=\dfrac{2+\sqrt{2}}{4}\\ \\ \cos \left(\dfrac{\pi}{8} \right )=\pm \sqrt{\dfrac{2+\sqrt{2}}{4}}\\ \\ \cos \left(\dfrac{\pi}{8} \right )=\pm \dfrac{\sqrt{2+\sqrt{2}}}{2}$


Como $\dfrac{\pi}{8}$ é um arco do primeiro quadrante, o seu cosseno também é positivo. Logo,

$\cos \left(\dfrac{\pi}{8} \right )=\dfrac{\sqrt{2+\sqrt{2}}}{2}$


Encontrando a tangente,

$\bullet\;\; \mathrm{tg}\left(\dfrac{\pi}{8} \right )=\dfrac{\mathrm{sen\frac{\pi}{8}}}{\cos \frac{\pi}{8}}\\ \\ \\ \mathrm{tg}\left(\dfrac{\pi}{8} \right )=\dfrac{\frac{\sqrt{2-\sqrt{2}}}{2}}{\frac{\sqrt{2+\sqrt{2}}}{2}}\\ \\ \\ \mathrm{tg}\left(\dfrac{\pi}{8} \right )=\dfrac{\sqrt{2-\sqrt{2}}}{\diagup\!\!\!\! 2}\cdot \dfrac{\diagup\!\!\!\! 2}{\sqrt{2+\sqrt{2}}}\\ \\ \\ \mathrm{tg}\left(\dfrac{\pi}{8}\right )=\dfrac{\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}\\ \\ \\ \mathrm{tg}\left(\dfrac{\pi}{8} \right )=\sqrt{\dfrac{2-\sqrt{2}}{2+\sqrt{2}}}\\ \\ \\ \mathrm{tg}\left(\dfrac{\pi}{8} \right )=\sqrt{\dfrac{\left(2-\sqrt{2} \right )\left(2-\sqrt{2} \right )}{\left(2+\sqrt{2} \right )\left(2-\sqrt{2} \right )}}\\ \\ \\ \mathrm{tg}\left(\dfrac{\pi}{8} \right )=\sqrt{\dfrac{\left(2-\sqrt{2} \right )^{2}}{4-2}}$


$\mathrm{tg}\left(\dfrac{\pi}{8} \right )=\sqrt{\dfrac{\left(2-\sqrt{2} \right )^{2}}{2}}\\ \\ \\ \mathrm{tg}\left(\dfrac{\pi}{8} \right )=\dfrac{\sqrt{\left(2-\sqrt{2} \right )^{2}}}{\sqrt{2}}\\ \\ \\ \mathrm{tg}\left(\dfrac{\pi}{8} \right )=\dfrac{2-\sqrt{2}}{\sqrt{2}}\\ \\ \\ \mathrm{tg}\left(\dfrac{\pi}{8} \right )=\dfrac{\left(2-\sqrt{2} \right )\cdot \sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}\\ \\ \\ \mathrm{tg}\left(\dfrac{\pi}{8} \right )=\dfrac{2\sqrt{2}-2}{2}\\ \\ \\ \mathrm{tg}\left(\dfrac{\pi}{8} \right )=\dfrac{\diagup\!\!\!\! 2\left(\sqrt{2}-1 \right )}{\diagup\!\!\!\! 2}\\ \\ \\ \mathrm{tg}\left(\dfrac{\pi}{8} \right )=\sqrt{2}-1$