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Math Help - A little help with de Moivre's

  1. #1
    Super Member craig's Avatar
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    A little help with de Moivre's

    Hi, could someone just give me a pointer where to start on this question.

    Use de Moivre's Theorem to show that \sin{5\theta} = 16\sin^5{\theta} - 20\sin^3{\theta} + 5\sin{\theta}.

    I know that de Moivre's Theorem states that (\cos{\theta} + i\sin{\theta})^n =  \cos{n\theta} + i\sin{n\theta}, just not sure where to start seeing as the question contains no \cos or no i?

    Thanks in advance
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    Quote Originally Posted by craig View Post
    Hi, could someone just give me a pointer where to start on this question.

    Use de Moivre's Theorem to show that \sin{5\theta} = 16\sin^5{\theta} - 20\sin^3{\theta} + 5\sin{\theta}.

    I know that de Moivre's Theorem states that (\cos{\theta} + i\sin{\theta})^n =  \cos{n\theta} + i\sin{n\theta}, just not sure where to start seeing as the question contains no \cos or no i?

    Thanks in advance
    Hi

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

    Expand the left hand side and take the imaginary part.
    Then substitute \cos^2 \theta by 1-\sin^2 \theta
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  3. #3
    Super Member craig's Avatar
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    Got it! Expand and equate the imaginary parts.

    Thanks for the reply
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