Question

calcule arc sen (cos 33pi/5)

Answer 1

Seja $\theta=\frac{33\pi}{5}.$ Então,

$\theta=\frac{3\pi+30\pi}{5}\\ \\ \theta=\frac{3\pi}{5}+\frac{30\pi}{5}\\ \\ \theta=\frac{3\pi}{5}+6\pi\\ \\ \theta=\frac{3\pi}{5}+3\cdot 2\pi\\ \\ \theta=\frac{3\pi}{5}+\frac{\pi}{2}-\frac{\pi}{2}+3\cdot 2\pi\\ \\ \theta=\frac{11\pi}{10}-\frac{\pi}{2}+3\cdot 2\pi$


Temos que,

$\cos \theta=\mathrm{sen}\left(\theta+\frac{\pi}{2}\right)\\ \\ \cos \frac{33\pi}{5}=\mathrm{sen}\left(\frac{33\pi}{2}+\frac{\pi}{2}\right)\\ \\ \cos \frac{33\pi}{5}=\mathrm{sen}\left(\frac{11\pi}{10}-\frac{\pi}{2}+3\cdot 2\pi+\frac{\pi}{2}\right)\\ \\ \cos \frac{33\pi}{5}=\mathrm{sen\,}\frac{11\pi}{10}\\ \\ \cos \frac{33\pi}{5}=\mathrm{sen}\left(\pi-\frac{11\pi}{10}\right )\\ \\ \cos \frac{33\pi}{5}=\mathrm{sen}\left(-\frac{\pi}{10}\right )$


Todas essas manipulações foram feitas de forma que o arco da função seno esteja no intervalo $\left[-\frac{\pi}{2};\,\frac{\pi}{2} \right ],$ pois esta é a imagem da função arco-seno.


Sendo assim, temos que

$\mathrm{arcsen}\left(\cos \frac{33\pi}{5}\right)\\ \\ =\mathrm{arcsen}\left[ \mathrm{sen}\left(-\frac{\pi}{10} \right )\right]\\ \\ =-\frac{\pi}{10}$