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Math Help - De Moivre's theorem

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    Newbie DefilerAlpha's Avatar
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    De Moivre's theorem

    Question: Use De Moivre's theorem to turn the functikons of the multiple angle into a polynomial in the sine or cosine of the angle theta as indicated.

    (a) cos(6theta) in terms of cos(theta)

    I have no idea at all where to begin on this, my lecture notes are unhelpful... any help would be most appreciated thanks!
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    Moo
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    Hello !

    Quote Originally Posted by DefilerAlpha View Post
    Question: Use De Moivre's theorem to turn the functikons of the multiple angle into a polynomial in the sine or cosine of the angle theta as indicated.

    (a) cos(6theta) in terms of cos(theta)

    I have no idea at all where to begin on this, my lecture notes are unhelpful... any help would be most appreciated thanks!
    De Moivre's theorem states :

    (\cos x+i \sin x)^n=\cos nx+i \sin nx

    --> \cos nx=\Re (\cos nx+i \sin nx)=\Re (\cos x+i \sin x)^n \qquad \Re \ : \ \text{Real part}

    \implies \cos 6 \theta=\Re (\cos \theta+i \sin \theta)^6

    Now use Newton's Binomial formula to get it...
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  3. #3
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    Quote Originally Posted by DefilerAlpha View Post
    Question: Use De Moivre's theorem to turn the functikons of the multiple angle into a polynomial in the sine or cosine of the angle theta as indicated.

    (a) cos(6theta) in terms of cos(theta)

    I have no idea at all where to begin on this, my lecture notes are unhelpful... any help would be most appreciated thanks!
    His theorem states:

    [\cos(\theta)+i\sin(\theta)]^n=\cos(n\theta)+i\sin(n\theta)

    Using n=6 we get

    [\cos(\theta)+i\sin(\theta)]^6=\cos(6\theta)+i\sin(6\theta)

    Expanding the left hand side with the binomial theorem (you could do it by hand, but I'm lazy) we get

    \cos^6(\theta)+i\cos^5(\theta)\sin(\theta)-\cos^4(\theta)\sin^2(\theta)-i\cos^3(\theta)\sin^3(\theta)+
    \cos^2(\theta)\sin^4(\theta)+i\cos(\theta)\sin^5(\  theta)-\sin^6(\theta) =\cos(6\theta)+i\sin(6\theta)

    Now we can equate the real part of both sides of the equation to get

    \cos^{6}(\theta)-\cos^{4}(\theta)\sin^2(\theta)+\cos^2(\theta)\sin^  4(\theta)-\sin^6(\theta)=\cos(6\theta)

    Yeah!!
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