# principle root and cosines.

• Mar 18th 2007, 10:21 AM
principle root and cosines.
Hi MHF,

I needed help on two question please, that are:

1] Find the three cube roots of 8(cos 264º + j sin 264º) and state which of them is the principal cube root. Show all three roots on an Argand diagram.

2] i) Expand sin 4θ in powers of sinθ and cosθ.
ii) Expand cos^4θ in terms of cosines of multiples of θ.

Thank you.
• Mar 18th 2007, 11:52 AM
CaptainBlack
Quote:

Hi MHF,

I needed help on two question please, that are:

1] Find the three cube roots of 8(cos 264º + j sin 264º) and state which of them is the principal cube root. Show all three roots on an Argand diagram.

8(cos 264º + j sin 264º) = 8 cis(264 + 360*n), n=0, +/-1, +/-2, ..

So:

(8(cos 264º + j sin 264º))^{1/3} = 2 cis(264/3 + 360*n/3) n =...

............= 2 cis(88 + 120*n) n=0, +/-1, +/-2, ..

Now only three of these are distinct, and taking n=0, -1, +1 will give us
a complete set.

These are:

2 cis(88), 2 cis(-32), 2 cis(208).

The principal root is the one with positive imaginary part, which is 8 cis(88).

RonL
• Mar 18th 2007, 12:08 PM
CaptainBlack
Quote:

2] i) Expand sin 4θ in powers of sinθ and cosθ.

cis(4 theta) = [cis(theta)]^4 = [cos(theta) + i sin(theta)]^4

.................= c^4 + 4 c^3 (i s) + 6 c^2 (i s)^2 + 4 c (i s)^3 + (i s)^4

.................= c^4 + 4 i c^3 s - 6 c^2 s^2 -i 4 c s^3 + s^4

where c denites cos(theta) and s sin(theta).

Now sin(4 theta) = Im (cis(4 theta)), so is equal to:

4 cos^3(theta) sin(theta) - 4 cos(theta) sin^3(theta).

RonL
• Mar 18th 2007, 03:11 PM
Soroban

For #2, you need some identities:

. . sin(2x) . = . 2·sin(x)·cos(x)
. . cos(2x) . = . cos²x - sin²x . = . 2·cos²x - 1 . = . 1 - 2·sin²x
. . cos²(x) . = . ½[1 + cos(2x)]

Quote:

2] a) Expand sin(4θ) in powers of sin(θ) and cos(θ).
We have: . sin(4θ) . = . 2·sin(2θ)·cos(2θ)

. . . . . . . . . . . . . . .= . 2 [2·sin(θ)·cos(θ)]·[cos²(θ) - sin²(θ)]

. . . . . . . . . . . . . . .= . 4·sin(θ)·cos³(θ) - 4·sin³(θ)·cos(θ)

Quote:

2] b) Expand cos^4(θ) in terms of cosines of multiples of θ.

cos^4(T) . = . [cos²(θ)]² . = . [½(1 + cos(2θ)]²

. . . . . . . .= . [1 + 2·cos(2θ) + cos²(2θ)]/4 . = . [1 + 2·cos(2θ) + ½{1 + cos(4θ)}]/4

. . . . . . . .= . [2 + 4·cos(2θ) + 1 + cos(4θ)]/8 . = . [3 + 4·cos(2θ) + cos(4θ)]/8