Question

Please helpp me?!!!! I will fail my Calc class!!

Answer 1

Step-by-step explanation:

Using that fact that,

$\cos(u) = \frac{x}{r}$

and

${r}^{2} = {x}^{2} + {y}^{2}$

We need to find sin which that formua is

$\sin(u) = \frac{y}{r}$

We know what r and x is so we need to find y.

${8}^{2} = { - 3}^{2} + {y}^{2}$

$64 = 9 + {y}^{2}$

$55 = {y}^{2}$

$\sqrt{55} = y$

Sin is positve on the interval pi/2 to pi. so we get

$\sin(u) = \frac{ \sqrt{55} }{8}$

b.

$\tan(u) = \frac{y}{x}$

so

$\tan(u) = - \frac{ \sqrt{55} }{3}$

c.

$\sec(u) = \frac{r}{ x}$

$\sec(u) = - \frac{8}{3}$

d.

$\sin(2u) = 2 \sin(u) \cos(u)$

$2 ( \frac{ \sqrt{55} }{8} )( - \frac{3}{8} ) = \frac{ - 6 \sqrt{55} }{64} = \frac{ - 3 \sqrt{55} }{32}$

e.

$\cos {}^{} (2u) = \cos {}^{2} (u) - \sin {}^{2} (u)$

$( \frac{ - 3}{8} ) {}^{2} - ( \frac{ \sqrt{55} }{8} ) {}^{2} = \frac{ - 46}{64} = - \frac{23}{32}$