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Math Help - Equation solving trigonomeric

  1. #1
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    Equation solving trigonomeric

    I have an exercise as follows:
    Show that: cos(5*angle)=16*cos^5(angle)-20*cos^3(angle)+cos(angle), and hence show that the roots of: x(16x^4-20x^2+5)=0 are 0; cos(pi/10); cos(3*pi/10); cos(7*pi/10) and cos(9*pi/10)

    The first part has me completely confused, but the second part is also confusing.

    Could you show me what way to do this?

    thx
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by johnnyboy View Post
    I have an exercise as follows:
    Show that: cos(5*angle)=16*cos^5(angle)-20*cos^3(angle)+cos(angle), and hence show that the roots of: x(16x^4-20x^2+5)=0 are 0; cos(pi/10); cos(3*pi/10); cos(7*pi/10) and cos(9*pi/10)

    The first part has me completely confused, but the second part is also confusing.

    Could you show me what way to do this?

    thx
    cos(5 theta)=Re[e^(5 i theta)]=Re [(e^(i theta))^5]

    ......=Re[ (cos(theta)+isin(theta))^5]

    ......=Re[c^5+5 i s c^4-10 s^2 c^3 -10 i s^3 c^2 +5 s^4 c + i s^5]

    ......=c^5 - 10(1-c^2)c^3 + 5(1-c^2)^2 c

    ......=16 c^5 -20 c^3 + 5 c

    Thence the roots of:

    x(16x^4-20x^2+5)=0

    are the roots of cos(5 x)=0, and the given roots are distinct roots of this
    equation.

    RonL
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  3. #3
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    Hello, Johnny!

    Here's the first part . . .

    You're expected to know these identities:

    . . cos(A + B) .= .cos(A)Ěcos(B) - sin(A)Ěsin(B)

    . . sin(2A) .= .2Ěsin(A)Ěcos(A)

    . . cos(2A) .= .2Ěcos2(A) - 1


    Show that: .cos(5θ) .= .16Ěcos^5(θ) - 20Ěcos│(θ) + cos(θ)

    cos(5θ) .= .cos(4θ + θ)

    . . . . . . = .cos(4θ)cos(θ) - sin(4θ)sin(θ)

    . . . . . . = .[2cos▓(2θ) - 1]Ěcos(θ) - [2Ěsin(2θ)Ěcos(2θ)]Ěsin(θ)

    . . . . . . = .2Ěcos▓(2θ)Ěcos(θ) - cos(θ) - 2Ě[2Ěsin(θ)Ěcos(θ)][2Ěcos▓(θ) - 1]Ěsin(θ)

    . . . . . . = .2Ě[2Ěcos▓(θ) - 1]▓Ěcos(θ) - cos(θ) - 4Ěsin▓(θ)Ěcos(θ)[2Ěcos▓(θ) - 1]

    . . . . . . = .2Ě[4Ěcos^4(θ) - 4Ěcos▓(θ) + 1]Ěcos(θ) - cos(θ) - 4Ě[1 - cos▓(θ)]Ěcos(θ)Ě[2Ěcos▓(θ) - 1]

    . . . . . . = .8Ěcos^5(θ) - 8Ěcos│(θ) + 2Ěcos(θ) - cos(θ) - 4Ěcos(θ)&#183l[1 - cos▓(θ)][2Ěcos▓(θ) - 1]

    . . . . . . = .8Ěcos^5(θ) - 8Ěcos│(θ) + cos(θ) - 8Ěcos│(θ) + 8Ěcos^5(θ) + 4Ěcos(θ) - 4Ěcos│(θ)

    . . . . . . = .16Ěcos^5(θ) - 20Ěcos│(θ) + 5Ěcos(θ)

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Captain Black's method is much simpler
    . . if you're familiar with DeMoivre's Theorem.


    Edit: Corrected . . . You're right, Captain, it's a 5.
    Last edited by Soroban; December 23rd 2006 at 12:40 PM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Soroban View Post
    Hello, Johnny!

    Here's the first part . . .

    You're expected to know these identities:

    . . cos(A + B) .= .cos(A)Ěcos(B) - sin(A)Ěsin(B)

    . . sin(2A) .= .2Ěsin(A)Ěcos(A)

    . . cos(2A) .= .2Ěcos2(A) - 1


    cos(5θ) .= .cos(4θ + θ)

    . . . . . . = .cos(4θ)cos(θ) - sin(4θ)sin(θ)

    . . . . . . = .[2cos▓(2θ) - 1]Ěcos(θ) - [2Ěsin(2θ)Ěcos(2θ)]Ěsin(θ)

    . . . . . . = .2Ěcos▓(2θ)Ěcos(θ) - cos(θ) - 2Ě[2Ěsin(θ)Ěcos(θ)][2Ěcos▓(θ) - 1]Ěsin(θ)

    . . . . . . = .2Ě[2Ěcos▓(θ) - 1]▓Ěcos(θ) - cos(θ) - 4Ěsin▓(θ)Ěcos(θ)[2Ěcos▓(θ) - 1]

    . . . . . . = .2Ě[4Ěcos^4(θ) - 4Ěcos▓(θ) - 1]Ěcos(θ) - cos(θ) - 4Ě[1 - cos▓(θ)]Ěcos(θ)Ě[2Ěcos▓(θ) - 1]

    . . . . . . = .8Ěcos^5(θ) - 8Ěcos│(θ) - 2Ěcos(θ) - cos(θ) - 4Ěcos(θ)&#183l[1 - cos▓(θ)][2Ěcos▓(θ) - 1]

    . . . . . . = .8Ěcos^5(θ) - 8Ěcos│(θ) - 3Ěcos(θ) - 8Ěcos│(θ) + 8Ěcos^5(θ) + 4Ěcos(θ) - 4Ěcos│(θ)

    . . . . . . = .16Ěcos^5(θ) - 20Ěcos│(θ) + cos(θ)

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Captain Black's method is much simpler
    . . if you're familiar with DeMoivre's Theorem.
    We seem to disagree about the coefficient of cos. Yours agrees with
    that in the question, but I had assumed that was a typo, because of
    the second part of the question.

    I have just checked with a symbolic maths package and it gives 5 as
    this coefficient. There can't be something interesting going on here,
    can there?

    RonL
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  5. #5
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    Thanks for the help guys.
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