Question

Calcule o valor do seno e tangente dos ângulos abaixo, sabendo que:a) cos$\alpha = \frac{1}{2}$b) cos$\alpha = \frac{3}{5}$c) cos$\alpha = \sqrt{2}/2$

Answer 1

$a)Sen^2(x) + ( \frac{1}{2})^2 = 1 \\ Sen^2(x) = 1- \frac{1}{4} \\ Sen(x) = \sqrt{ \frac{3}{4} } = \frac{ \sqrt{3} }{2}$

$a) Tg(x) = \frac{Sen(x)}{Cos(x)} \\ \\ Tg(x) = \frac{ \frac{ \sqrt{3} }{2} }{ \frac{1}{2} } \\ \\ Tg(x) = \frac{ \sqrt{3} }{2} } . \frac{2}{1} = \sqrt{3}$

$a)[tex]b)[tex]Sen^2(x) + Cos^2(x) = 1 \\ \\ Sen^2(x) + ( \frac{ \sqrt{2} }{2}) ^2 = 1 \\ \\ Sen^2(x) = 1 - \frac{1}{2} = \frac{1}{2} \\ \\ Sen(x) = \frac{ \sqrt{1} }{ \sqrt{2} } = \frac{1}{ \sqrt{2} } = \frac{ \sqrt{2} }{2}$

$[tex]c)Tg(x) = \frac{sen(x)}{cos(x)} \\ \\ Tg(x) = \frac{ \frac{ \sqrt{2} }{2} }{ \frac{ \sqrt{2} }{2} } = \frac{ \sqrt{2} }{2} . \frac{2}{ \sqrt{2} } = 1$ [/tex]