De Moivre's Theorem

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  1. Fing4's Avatar
    1 13-07-2012  13:51
    • Exalted Member
    • Posts: 255
    De Moivre's Theorem
    Hello, could somebody please point out where I am going wrong please?

    Question is show that sin6\theta=sin\theta(32cos^5 \theta-32cos^3\theta+6cos\theta) .

    I started out by letting z=cos\theta+isin\theta and trying (z+\dfrac{1}{z})^6 and (z-\dfrac {1}{z})^6 with no further progress as I only get cos(6theta) instead of sin.
    Last edited by Fing4; 13-07-2012 at 13:54.
  2. dknt's Avatar
    2 13-07-2012  13:58
    • Overlord in Training
    • Location: UK
    • Posts: 2,233
    Re: De Moivre's Theorem
    (Original post by Fing4)
    Hello, could somebody please point out where I am going wrong please?

    Question is show that sin6\theta=sin\theta(32cos^5 \theta-32cos^3\theta+6cos\theta) .

    I started out by letting z=cos\theta+isin\theta and trying (z+\dfrac{1}{z})^6 and (z-\dfrac {1}{z})^6 with no further progress as I only get cos(6theta) instead of sin.
    You'll get your \mathrm{sin}6\theta from (\mathrm{cos} \theta + i \mathrm{sin}\theta)^6 , using de Moivre's theorem. You can also then expand it binomially and equate imaginary parts. Then it's just a case of some trig identities.
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  3. Fing4's Avatar
    3 13-07-2012  14:07
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    • Posts: 255
    Re: De Moivre's Theorem
    (Original post by dknt)
    You'll get your \mathrm{sin}6\theta from (\mathrm{cos} \theta + i \mathrm{sin}\theta)^6 , using de Moivre's theorem. You can also then expand it binomially and equate imaginary parts. Then it's just a case of some trig identities.
    The previous question was to find sin5\theta and I did it by setting (2isin\theta)^5=(z-\dfrac{1}{z})^5 and using trig identities for sin3\theta .

    Is there a reason why this method didn't work for sin6\theta ?

    Edit: Oh and thank you! Much easier using the 'conventional' de Moivre's theorem way rather than using the z+1/z stuff
    Last edited by Fing4; 13-07-2012 at 14:15.
  4. dknt's Avatar
    4 13-07-2012  14:30
    • Overlord in Training
    • Location: UK
    • Posts: 2,233
    Re: De Moivre's Theorem
    (Original post by Fing4)
    The previous question was to find sin5\theta and I did it by setting (2isin\theta)^5=(z-\dfrac{1}{z})^5 and using trig identities for sin3\theta .

    Is there a reason why this method didn't work for sin6\theta ?

    Edit: Oh and thank you! Much easier using the 'conventional' de Moivre's theorem way rather than using the z+1/z stuff
    Well it should work both ways whether you start on the LHS and work to the RHS or from RHS to LHS, as its an identity, but one way will require a lot more work. For example you'd have to use a lot more identities and would probably just end up with even more (confusing) steps.

    As a note, if you're asked to show for example \mathrm{sin} nx = terms in \mathrm{sin}^ax/\mathrm{cos}^ax use the method I just used. However there may be times when they ask you to show for example \mathrm{sin}^n x = terms in \mathrm{sin}a x/ \mathrm{cos} ax, in which case use the (z -1/z) etc and expand out etc. Probably best to do an example yourself.
    Last edited by dknt; 13-07-2012 at 14:53.
  5. Fing4's Avatar
    5 13-07-2012  14:38
    • Exalted Member
    • Posts: 255
    Re: De Moivre's Theorem
    Thanks!
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