Question

Trigonometria...
Sabendo-se que x + y = Pi/4 e sen(x) = 5/13, com 0 < x < Pi/2. Calcule sen(y) e cos(y).

Answer 1

Primeiramente, para usar as relações trigonométrica de soma de arcos, primeiro temos que saber o seno e o cosseno de x. O seno já é conhecido, é preciso saber o cosseno

$\mathsf{sen^{2}x+ cos^{2}x=1}\\\\\\ \mathsf{\left(\dfrac{5}{13}\right)^{2}+ cos^{2}x=1}\\\\\\ \mathsf{cos^{2}x=1-\left(\dfrac{5}{13}\right)^{2}}\\\\\\ \mathsf{cos^{2}x=1- \dfrac{25}{169}}\\\\\\ \mathsf{cos^{2}x=\dfrac{144}{169}}\\\\\\ \mathsf{cos\ x= \sqrt{\dfrac{144}{169}}}\\\\\\ \mathsf{cos\ x=\dfrac{12}{13}}$


ok

__________


Isolando o y na sentença que o exercício cedeu obtemos

$\mathsf{x+y=\dfrac{\pi}{4}}\\\\\\ \mathsf{y=\dfrac{\pi}{4}-x}$


Logo, podemos afirmar que

$\mathsf{sen\ y=sen\ \left(\dfrac{\pi}{4}-x\right)}$


Desta forma podemos obter sen y com a relação trigonométrica da soma de arcos

$\mathsf{sen\ \left(\dfrac{\pi}{4}-x\right)=sen\ \dfrac{\pi}{4}\cdot cos\ x-sen\ x\cdot cos\ \dfrac{\pi}{4}}}\\\\\\ \mathsf{sen\ \left(\dfrac{\pi}{4}-x\right)=\dfrac{ \sqrt{2}}{2}\ \cdot \ \dfrac{12}{13}\ -\ \dfrac{5}{13}\ \cdot \ \dfrac{ \sqrt{2} }{2}}}\\\\\\ \mathsf{sen\ \left(\dfrac{\pi}{4}-x\right)=\dfrac{12\cdot \sqrt{2}}{26}\ - \ \dfrac{5\cdot \sqrt{2} }{26}}}\\\\\\ \mathsf{sen\ \left(\dfrac{\pi}{4}-x\right)=\dfrac{7\cdot \sqrt{2}}{26}}}$


$\boxed{\begin{array}{c} \mathsf{sen\ y=\dfrac{7\cdot \sqrt{2}}{26}} \end{array}}}$



Para o cosseno fazemos praticamente a mesma coisa, porém com a relação de soma de arcos de cosseno

$\mathsf{cos\ y=cos\ \left(\dfrac{\pi}{4}-x\right)}$


$\mathsf{cos\ \left(\dfrac{\pi}{4}-x\right)=cos\ \dfrac{\pi}{4}\cdot cos\ x+sen\ x\cdot sen\ \dfrac{\pi}{4}}}\\\\\\ \mathsf{cos\ \left(\dfrac{\pi}{4}-x\right)=\dfrac{ \sqrt{2} }{2}\ \cdot\ \dfrac{12}{13}\ +\ \dfrac{5}{13}\ \cdot\ \dfrac{ \sqrt{2} }{2} }\\\\\\ \mathsf{cos\ \left(\dfrac{\pi}{4}-x\right)=\dfrac{12\cdot \sqrt{2}}{26}\ + \ \dfrac{5\cdot \sqrt{2} }{26}}}\\\\\\ \mathsf{sen\ \left(\dfrac{\pi}{4}-x\right)=\dfrac{7\cdot \sqrt{2}}{26}}$


$\boxed{\begin{array}{c} \mathsf{cos\ y=\dfrac{17\cdot \sqrt{2}}{26}} \end{array}}}$


Bons estudos! :)