Question

look at attachment

Answer 1

Well, do you know the formula?

Answer 2

de Moivre's formula that is.

Answer 3

yes

Answer 4

Great, so you can apply that formula here to get the answer.

Answer 5

so B

Answer 6

No, the exponent, 4, goes into the sine and cosine.

Answer 7

oh A!

Answer 8

That's not what I meant, it's D. You multiply the input of the sine and cosine by 4. That's what de Moivre's formula states.

Answer 9

[e^(ix)]^n = (cosx + i.sinx)^n

But since (e^x)^n = e^(x×n) then

e^(i.nx) = (cosx + i.sinx)^n

But by Euler's original formula,

e^(i.nx) = cos(nx) + i.sin(nx), and therefore

(cosx + i.sinx)^n = cos(nx) + i.sin(nx)