Question

31. Calcule: b) sen a, sabendo que cos a = - 8/17, π < a < 3π/2 .

Answer 1

$\displaystyle \sf cos\ a = \frac{-8}{17} \ \ ;\ \ \pi < a < \frac{3\pi}{2} \\\\\ Pela\ rela\c c \~ao \ fundamental\ da \ trigonometria\ sabemos\ que : \\\\ sen^2a+cos^2a = 1 \\\\ sen^2a = 1-cos^2a \\\\ sen\ a=\pm\sqrt{1-cos^2a } \\\\ Substituindo\ o \ valor\ do\ cos\ a : \\\\ sen\ a=\pm \sqrt{1-\left(\frac{-8}{17}\right)} \\\\\\ sen\ a = \pm\sqrt{1-\frac{64}{289}} \\\\\\\ sen\ a = \pm\sqrt{\frac{289-64}{289}} \\\\\\ sen\ a =\pm\sqrt{\frac{225}{289}} \\\\\\ sen\ a = \pm\frac{15}{17}$

o ângulo está no 3º quadrante, logo o seno será negativo. Portanto :

$\huge\boxed{\displaystyle \sf sen\ a = \frac{-15}{17}\ }\checkmark$