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Sindrei [870]
3 years ago
6

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

Mathematics
1 answer:
expeople1 [14]3 years ago
5 0

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}

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