Step-by-step explanation:
Using that fact that,
[tex] \cos(u) = \frac{x}{r} [/tex]
and
[tex] {r}^{2} = {x}^{2} + {y}^{2} [/tex]
We need to find sin which that formua is
[tex] \sin(u) = \frac{y}{r} [/tex]
We know what r and x is so we need to find y.
[tex] {8}^{2} = { - 3}^{2} + {y}^{2} [/tex]
[tex]64 = 9 + {y}^{2} [/tex]
[tex]55 = {y}^{2} [/tex]
[tex] \sqrt{55} = y[/tex]
Sin is positve on the interval pi/2 to pi. so we get
[tex] \sin(u) = \frac{ \sqrt{55} }{8} [/tex]
b.
[tex] \tan(u) = \frac{y}{x} [/tex]
so
[tex] \tan(u) = - \frac{ \sqrt{55} }{3} [/tex]
c.
[tex] \sec(u) = \frac{r}{ x} [/tex]
[tex] \sec(u) = - \frac{8}{3} [/tex]
d.
[tex] \sin(2u) = 2 \sin(u) \cos(u) [/tex]
[tex]2 ( \frac{ \sqrt{55} }{8} )( - \frac{3}{8} ) = \frac{ - 6 \sqrt{55} }{64} = \frac{ - 3 \sqrt{55} }{32} [/tex]
e.
[tex] \cos {}^{} (2u) = \cos {}^{2} (u) - \sin {}^{2} (u) [/tex]
[tex]( \frac{ - 3}{8} ) {}^{2} - ( \frac{ \sqrt{55} }{8} ) {}^{2} = \frac{ - 46}{64} = - \frac{23}{32} [/tex]
