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    Find the general solution of:

     sin\theta + sin2\theta + sin3\theta = 0

    Unfortunately I wan't here for this part of the course so I am slightly confused. Could anyone please help?

    Do I have to use  t = tan(\frac{\theta}{2}) ?
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    No, you need to use De Moivre's theorem to write \sin 2\theta and \sin 3\theta in terms of \sin \theta and \cos \theta. Then factorise and solve for \theta. [EDIT: Mr M's method is a lot cleaner... spotting the GP definitely helps!]

    The substitution t=\tan\frac{\theta}{2} is used in integration problems, whereas this is an algebraic problem.
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    sin t + sin 2t + sin 3t = Im(e^it + e^2it +e^3it).

    Then sum the GP on the RHS.

    Alternatively (but probably not the expected method):

    sin 2t = 2 sin t cos t. sin 3t = 3 sin t -4 sin^3t = sin t (3-4 sin^2 t) = sin t(3 - (4-4cos^t)) = sin t (4 cos^2 t - 3).

    Apply this to write sin t + sin 2t + sin 3t = (sin t) P, where P is a polynomial in cos t. Then fully factor the RHS.
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    just use sinA+sinB=2sin(A+B/2)cos(A-B/2) ..for sin3 \theta and sin \theta
    or

    Im[e^{ix}+{e^{2ix}}+{e^{3ix}}] notice is a GP
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    (Original post by rbnhelp)
    ..
    I don't think you meant your LaTeX to look lilke that.
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    (Original post by nuodai)
    No, you need to use De Moivre's theorem to write \sin 2\theta and \sin 3\theta in terms of \sin \theta and \cos \theta. Then factorise and solve for \theta. [EDIT: Mr M's method is a lot cleaner... spotting the GP definitely helps!]

    The substitution t=\tan\frac{\theta}{2} is used in integration problems, whereas this is an algebraic problem.
    Okay, it was just I have seen a similar problem solved in that way.
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    (Original post by DFranklin)
    sin t + sin 2t + sin 3t = Im(e^it + e^2it +e^3it).

    Then sum the GP on the RHS.


    Alternatively (but probably not the expected method):

    sin 2t = 2 sin t cos t. sin 3t = 3 sin t -4 sin^3t = sin t (3-4 sin^2 t) = sin t(3 - (4-4cos^t)) = sin t (4 cos^2 t - 3).

    Apply this to write sin t + sin 2t + sin 3t = (sin t) P, where P is a polynomial in cos t. Then fully factor the RHS.
    Could I ask what chapter that part is in? Because I'm not completely sure if it's in my curriculum. How do you get e^something from sin3t?
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    (Original post by gunmetalpanda)
    Could I ask what chapter that part is in? Because I'm not completely sure if it's in my curriculum. How do you get e^something from sin3t?
    FP2.
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    (Original post by gunmetalpanda)
    Could I ask what chapter that part is in? Because I'm not completely sure if it's in my curriculum. How do you get e^something from sin3t?
    Have you covered De Moivre's theorem before (complex numbers)? It's usually either in FP2 or FP3.
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    (Original post by StephenP91)
    FP2.
    Not on every specification. Definitely not on OCR's.
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    (Original post by Farhan.Hanif93)
    Not on every specification. Definitely not on OCR's.
    Fair enough. I apologise. For WJEC it is FP2.
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    (Original post by gunmetalpanda)
    Could I ask what chapter that part is in? Because I'm not completely sure if it's in my curriculum. How do you get e^something from sin3t?
    FP2 or FP3, usually, depending on exam board.

    It comes from the identity e^{i\theta} \equiv \cos \theta + i \sin \theta, and hence that e^{in\theta} = \cos n \theta + i\sin n \theta = (\cos \theta + i\sin \theta)^n; it allows you to express trig functions in terms of exponentials by comparing real and imaginary parts.
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    (Original post by DFranklin)

    sin 2t = 2 sin t cos t. sin 3t = 3 sin t -4 sin^3t = sin t (3-4 sin^2 t) = sin t(3 - (4-4cos^t)) = sin t (4 cos^2 t - 3).

    Apply this to write sin t + sin 2t + sin 3t = (sin t) P, where P is a polynomial in cos t. Then fully factor the RHS.
    easily beat that ,

    (sinx+sin3x)+sin2x=0
    2sin2xcosx+sin2x=0
    sin2x=0,cosx=-1/2
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    (Original post by rbnphlp)
    easily beat that ,

    (sinx+sin3x)+sin2x=0
    2sin2xcosx+sin2x=0
    sin2x=0,cosx=-1/2
    And that, ladies and gentleman, is why you should remember the sum and product trig identities. (I don't, I'm afraid).
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    (Original post by DFranklin)
    And that, ladies and gentleman, is why you should remember the sum and product trig identities. (I don't, I'm afraid).
    I don't know those off the top of my head but I do know the Chebyshev method which is equally quick:

    \sin nx = 2 \cos x \sin (n-1) x - \sin (n-2) x

    \sin 3x = 2 \cos x \sin 2x - \sin x
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    (Original post by Mr M)
    I don't know those off the top of my head but I do know the Chebyshev method which is equally quick:

    \sin nx = 2 \cos x \sin (n-1) x - \sin (n-2) x

    \sin 3x = 2 \cos x \sin 2x - \sin x
    I've never come across that before. Definitely looks like it's worth remembering!
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    (Original post by Farhan.Hanif93)
    I've never come across that before. Definitely looks like it's worth remembering!
    Easy to remember as the cos version is so similar:

    \cos nx = 2 \cos x \cos (n-1)x - \cos (n-2) x

    Now you can spend a few minutes proving them!
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    (Original post by rbnphlp)
    easily beat that ,

    (sinx+sin3x)+sin2x=0
    2sin2xcosx+sin2x=0
    sin2x=0,cosx=-1/2
    This was the first thing that occurred to me, but what makes it FP2?
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    (Original post by tiny hobbit)
    This was the first thing that occurred to me, but what makes it FP2?
    The (probably intended) use of De Moivre's, I'd imagine but I agree that it's easier solved by more basic methods.
 
 
 
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