Question

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

Answer 1

Explanation:

Using that fact that,

`\cos(u) = (x)/(r)`

and

`{r}^(2) = {x}^(2) + {y}^(2)`

We need to find sin which that formua is

`\sin(u) = (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)`

`√(55) = y`

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

`\sin(u) = ( √(55) )/(8)`

b.

`\tan(u) = (y)/(x)`

so

`\tan(u) = - ( √(55) )/(3)`

c.

`\sec(u) = (r)/( x)`

`\sec(u) = - (8)/(3)`

d.

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

`2 ( ( √(55) )/(8) )( - (3)/(8) ) = ( - 6 √(55) )/(64) = ( - 3 √(55) )/(32)`

e.

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

`( ( - 3)/(8) ) {}^(2) - ( ( √(55) )/(8) ) {}^(2) = ( - 46)/(64) = - (23)/(32)`