Question

dado que seno de x mais cosseno de x é igual a um dividido por 2 calcule seno de x e cosseno de x

Answer 1

As identidades utilizadas para resolver esta equação se encontram no arquivo em anexo.


Resolver a equação:

$\mathrm{sen\,}x+\cos x=\frac{1}{2}\\ \\ \mathrm{sen\,}x+\mathrm{sen}\left(\,^{\pi}\!\!\!\diagup \!\!_{2} -x\right )=\frac{1}{2}$


Utilizando a identidade $\text{(ii)}$, no lado esquerdo da equação acima, para

$\alpha=x,\;\;\;\beta=\,^{\pi}\!\!\!\diagup\!\!_{2} -x$

temos


$2\cdot\mathrm{sen}\left(\frac{x+\left(\,^{\pi}\!\!\!\diagup\!\!_{2} -x \right )}{2} \right )\cdot \cos\left(\frac{x-\left(\,^{\pi}\!\!\!\diagup\!\!_{2} -x \right )}{2} \right )=\frac{1}{2}\\ \\ 2\cdot\mathrm{sen}\left(\frac{\,^{\pi}\!\!\!\diagup\!\!_{2}}{2} \right )\cdot \cos\left(\frac{x-\,^ {\pi}\!\!\!\diagup\!\!_{2}+x}{2} \right )=\frac{1}{2}\\ \\ 2\cdot\mathrm{sen}\left(\,^{\pi}\!\!\! \diagup\!\!_{4}\right)\cdot \cos\left(\frac{2x-\,^{\pi}\!\!\!\diagup\!\!_{2}}{2} \right )=\frac{1} {2}\\ \\ 2\cdot\mathrm{sen}\left(\,^{\pi}\!\!\!\diagup\!\!_{4}\right)\cdot \cos\left(\frac{2x}{2}- \frac{\,^{\pi}\!\!\!\diagup\!\!_{2}}{2} \right )=\frac{1}{2}\\ \\ 2\cdot\frac{\sqrt{2}}{2}\cdot \cos \left(x-\,^{\pi}\!\!\!\diagup\!\!_{4}\right )=\frac{1}{2}\\ \\ \sqrt{2}\cdot\cos\left(x-\,^{\pi}\!\! \!\diagup\!\!_{4}\right )=\frac{1}{2}\\ \\ \cos\left(x-\,^{\pi}\!\!\!\diagup\!\!_{4}\right )=\frac{1} {2\sqrt{2}}$

$\cos\left(x-\,^ {\pi}\!\!\!\diagup\!\!_{4}\right )=\frac{1 \cdot \sqrt{2}}{2\sqrt{2} \cdot \sqrt{2}}\\ \\ \cos\left (x-\,^{\pi}\!\!\!\diagup\!\!_{4}\right )=\frac{\sqrt{2}}{4}\\ \\ x-\frac{\pi}{4}=\pm \arccos\left(\,^ {\sqrt{2}}\!\!\!\diagup\!\!_{4} \right )+k \cdot 2\pi\\ \\ x= \frac{\pi}{4}\pm \arccos\left(\,^{\sqrt {2}}\!\!\!\diagup\!\!_{4} \right )+k \cdot 2\pi\\ \\ x= \frac{\pi}{4}\pm \theta+k \cdot 2\pi\\ \\ \boxed{ \begin{array}{rcl} x=\frac{\pi}{4}+ \theta +k \cdot 2\pi&\text{ ou }&x=\frac{\pi}{4}-\theta+k \cdot 2\pi \end{array} }$


onde 

$\theta=\arccos\left(\,^{\sqrt{2}}\!\!\!\diagup\!\!_{4} \right )\\ \\ \cos \theta=\frac{\sqrt {2}}{4},\;\;\;\mathrm{sen\,}\theta=\frac{\sqrt{14}}{4}$

(observe o triângulo retângulo da figura em anexo)


e $k$ pode assumir apenas valores inteiros.


Vamos calcular os valores de $\mathrm{sen\,}x$ e $\cos x$. Desprezamos a parcela $k \cdot 2\pi$ para os valores de $x$, pois o seno e o cosseno são periódicos e seus valores não se alteram se adicionarmos ao ângulo, múltiplos inteiros de $2\pi$. Então,


$\bullet\;\;$ para $x=\frac{\pi}{4}+\theta$, temos

$\mathrm{sen\,}x=\mathrm{sen}\left(\,^{\pi}\!\!\!\diagup\!\!_{4}+\theta \right )\\ \\ \mathrm{sen \,}x=\mathrm{sen}\left(\,^{\pi}\!\!\!\diagup\!\!_{4}\right )\cdot\cos \theta +\mathrm{sen\,}\theta \cdot\cos\left(\,^{\pi}\!\!\!\diagup\!\!_{4}\right )\\ \\ \mathrm{sen\,}x=\frac{\sqrt {2}}{2}\cdot \frac{\sqrt {2}}{4} +\frac{\sqrt {14}}{4} \cdot\frac{\sqrt {2}}{2}\\ \\ \mathrm{sen\,}x=\frac{2+\sqrt {28}}{8}\\ \\ \mathrm{sen\,}x=\frac{2+2\sqrt{7}}{8}\\ \\ \mathrm{sen\,}x=\frac{2\cdot\left(1+\sqrt{7} \right )}{2 \cdot 4}\\ \\ \boxed{\mathrm{sen\,}x=\frac{1+\sqrt{7}}{4}}$


$\cos x=\cos \left(\,^{\pi}\!\!\!\diagup\!\!_{4}+\theta \right )\\ \\ \cos x=\cos \left(\,^ {\pi}\!\!\!\diagup\!\!_{4} \right )\cdot \cos \theta - \mathrm{sen}\left(\,^{\pi}\!\!\!\diagup\!\!_{4} \right )\cdot \mathrm{sen\,}\theta\\ \\ \cos x=\frac{\sqrt {2}}{2}\cdot \frac{\sqrt {2}}{4} - \frac {\sqrt {2}}{2}\cdot \frac{\sqrt {14}}{2}\\ \\ \cos x=\frac{2-\sqrt{28}}{8}\\ \\ \cos x=\frac{2-2\sqrt {7}}{8}\\ \\ \cos x=\frac{2\cdot \left(1-\sqrt{7}\right)}{2 \cdot 4}\\ \\ \boxed{\cos x=\frac{1-\sqrt {7}}{4}}$


$\bullet\;\;$ para $x=\frac{\pi}{4}-\theta$, temos

$\mathrm{sen\,}x=\mathrm{sen}\left(\,^{\pi}\!\!\!\diagup\!\!_{4}-\theta \right )\\ \\ \mathrm{sen \,}x=\mathrm{sen}\left(\,^{\pi}\!\!\!\diagup\!\!_{4}\right )\cdot\cos \theta -\mathrm{sen\,}\theta \cdot\cos\left(\,^{\pi}\!\!\!\diagup\!\!_{4}\right )\\ \\ \mathrm{sen\,}x=\frac{\sqrt {2}}{2}\cdot \frac{\sqrt {2}}{4} -\frac{\sqrt {14}}{4} \cdot\frac{\sqrt {2}}{2}\\ \\ \mathrm{sen\,}x=\frac{2-\sqrt {28}}{8}\\ \\ \mathrm{sen\,}x=\frac{2-2\sqrt{7}}{8}\\ \\ \mathrm{sen\,}x=\frac{2\cdot\left(1-\sqrt{7} \right )}{2 \cdot 4}\\ \\ \boxed{\mathrm{sen\,}x=\frac{1-\sqrt{7}}{4}}$


$\cos x=\cos \left(\,^{\pi}\!\!\!\diagup\!\!_{4}-\theta \right )\\ \\ \cos x=\cos \left(\,^ {\pi}\!\!\!\diagup\!\!_{4} \right )\cdot \cos \theta + \mathrm{sen}\left(\,^{\pi}\!\!\!\diagup\!\!_{4} \right )\cdot \mathrm{sen\,}\theta\\ \\ \cos x=\frac{\sqrt {2}}{2}\cdot \frac{\sqrt {2}}{4} + \frac {\sqrt {2}}{2}\cdot \frac{\sqrt {14}}{2}\\ \\ \cos x=\frac{2+\sqrt{28}}{8}\\ \\ \cos x=\frac{2+2\sqrt {7}}{8}\\ \\ \cos x=\frac{2\cdot \left(1+\sqrt{7}\right)}{2 \cdot 4}\\ \\ \boxed{\cos x=\frac{1+\sqrt {7}}{4}}$