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Math Help - Pls solve this for me now.

  1. #1
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    Pls solve this for me now.

    Using De-Moivre’s Theory

    Find Cos5θ and Sin5θ



    I am finding it difficult to solve.
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  2. #2
    MHF Contributor red_dog's Avatar
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    By Moivre:

    (\cos\theta +i\sin\theta)^5=\cos 5\theta +i\sin 5\theta

    By Newton:

    (\cos\theta+i\sin\theta)^5=
    =\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta  +(5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta)i

    Then,

    \cos 5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta

    \sin 5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta
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  3. #3
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    Hints:
    e^{5i\theta }  = \left( {e^{i\theta } } \right)^5  = \left( {\cos (\theta ) + i\sin (\theta )} \right)^5  = \left( {\cos (5\theta ) + i\sin (5\theta )} \right)<br />
    \cos (5\theta ) = {\mathop{\rm Re}\nolimits} \left( {e^{5i\theta } } \right)
    {\mathop{\rm Re}\nolimits} \left( {\left[ {x + iy} \right]^5 } \right) = x^5  - 10x^3 y^2  + 5xy^4

    Let x = \cos (\theta )\,\& \,y = \sin (\theta )
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by golo102 View Post
    Using De-Moivre’s Theory

    Find Cos5θ and Sin5θ



    I am finding it difficult to solve.
    Funny. Jhevon gave you a hint in your other thread a while ago.

    Please don't double post. It wastes the helpers' time. See rule #1 here.

    -Dan
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