Question

Last 2

Answer 1

just the first line

Answer 2

would the 4 apply to the cos or isin?

Answer 3

4 is the r value, it has nothing to do with the cos and sin values

Answer 4

okay so

do I cube the entire equation?

Answer 5

you are given

[r(cos(theta) + isin(theta)]^n

identify your r, theta, and n values, then write your solution in the form

r^n * (cos(n*theta) + i sin(n*theta))

Answer 6

cos(7pi/9)+isin(7pi/9)^3

Answer 7

now identify what your r, n, and theta are

Answer 8

r=4
n=3
theta=7pi/9

Answer 9

good, so r^n * (cos(n*theta) + i sin(n*theta)) = ?

Answer 10

r^n * (cos(n*theta) + i sin(n*theta))
4^3*(cos(3*7pi/9)+isin(3*7pi/9))

Answer 11

good

it wants the solution in standard form so you just need to distribute the 4^3 and use your calculator to evaluate the costheta and sintheta values to get something in the form a + bi

Answer 12

64(1/2+(i)sqrt3/2)

Answer 13

good but distribute that 64

Answer 14

32+32isqrt3

Answer 15

good but as a minor change it wants the solution in a +bi form so

32 + 32sqrt(3)i

Answer 16

then go through the same steps for 17

Answer 17

okay same thing ofr the next

Answer 18

okay will be back in a few

Answer 19

cos(72)+isin(72)^5

r=1/2
n=5
theta=72

r^n * (cos(n*theta) + i sin(n*theta)) =
1/2^5*(cos(5*72)+isin(5*72))

1/32

Answer 20

good now convert to a + bi form

Answer 21

1+32i?

Answer 22

what are cos(5*72) and sin(5*72) = ?

Answer 23

1 and 0

Answer 24

good so putting it together gives us 1/32 + 0i