Hello, classicstrings!

You'rethat closein part (a) . . .

(a) Expand (cosθ +i·sinθ)^4, wherei² = -1.

Hence by using De Moivre's theorem, show that:

. . sin4θ/cos4θ .= .(4c³s - 4cs³) / (c^4 - 6c²s² + s^4), where c = cosθ, and s = sinθ

I expanded to get: c^4 + 4c³s·i- 6c²s² - 6cs³·i+ s^4,

and I see that: sin4θ/cos4θ is just s^4/c^4.

But not sure how to show the solution.

DeMoivre's Theorem says: .(cosθ +i·sinθ)^4 .= .cos4θ +i·sin4θ

Your expansion is: .cos4θ +i·sin4θ .= .(c^4 - 6c²s² + s^4) +i·(4c³s - 4cs³)

Equate real and imaginary components:

. . cos4θ .= .c^4 - 6c²s² + s^4

. . sin4θ .= .4c³s - 4cs³

. . . . . . . . . sin4θ . . . . . . . 4c³s - 4cs³

Therefore: . -------- . = . ----------------------

. . . . . . . . . cos4θ . . . . c^4 - 6c²s² + s^4