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Math Help - principle root and cosines.

  1. #1
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    Post 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.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by dadon View Post
    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
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by dadon View Post
    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
    Last edited by CaptainBlack; March 18th 2007 at 09:50 PM. Reason: correction of some errors
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  4. #4
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    Hello, dadon!

    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)]



    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)



    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

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