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  1. #1
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    complex numbers eulers formula

    Using eulers formula show that sin3x = 3sinx - 4sin^3x
    How do I do that?
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    Re: complex numbers eulers formula

    No need for Euler's formula.

    You know that \displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^n = \sum_{r = 0}^n {n\choose{r}}\cos^{n-r}{\theta}\sin^r{\theta} \end{align*} by the Binomial Theorem, and \displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^n = \cos{n\theta} + i\sin{n\theta} \end{align*} by DeMoivre's Theorem.

    Try applying each of these to \displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^3 \end{align*}, setting them equal, and equating real and imaginary parts
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    Re: complex numbers eulers formula

    Quote Originally Posted by Prove It View Post
    No need for Euler's formula.

    You know that \displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^n = \sum_{r = 0}^n {n\choose{r}}\cos^{n-r}{\theta}\sin^r{\theta} \end{align*} by the Binomial Theorem, and \displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^n = \cos{n\theta} + i\sin{n\theta} \end{align*} by DeMoivre's Theorem.

    Try applying each of these to \displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^3 \end{align*}, setting them equal, and equating real and imaginary parts
    Thats what I thought (personally, I prefer De Moivres method) but the question specifically stated to use the identites of cos and sin from eulers formula
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    Re: complex numbers eulers formula

    Hmm, well let's see...

    \displaystyle \begin{align*} e^{i\theta} &= \cos{\theta} + i\sin{\theta} \\ \\ e^{-i\theta} &= \cos{\theta} - i\sin{\theta} \\ \\ e^{i\theta} - e^{-i\theta} &= 2i\sin{\theta} \\ \sin{\theta} &= \frac{e^{i\theta} - e^{-i\theta}}{2i} \\ \\ \sin{3x} &= \frac{e^{3i\,x} - e^{-3i\,x}}{2i} \\ \\ 3\sin{x} - 4\sin^3{x} &= 3\left(\frac{e^{i\,x} - e^{-i\,x}}{2i}\right) - 4\left(\frac{e^{i\,x} - e^{-i\,x}}{2i}\right)^3 \\ &= \frac{3e^{i\,x} - 3e^{-i\,x}}{2i} - 4\left(\frac{e^{3i\,x} - 3e^{i\,x} + 3e^{-i\,x} - e^{-3i\,x} }{-8i}\right) \\ &= \frac{3e^{i\,x} - 3e^{-i\,x}}{2i} + \frac{e^{3i\,x} - 3e^{i\,x} + 3e^{-i\,x} - e^{-3i\,x}}{2i} \\ &= \frac{e^{3i\,x} - e^{-3i\,x}}{2i} \\ &= \sin{3x} \end{align*}

    Q.E.D.
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    Re: complex numbers eulers formula

    Quote Originally Posted by Prove It View Post
    Hmm, well let's see...

    \displaystyle \begin{align*} e^{i\theta} &= \cos{\theta} + i\sin{\theta} \\ \\ e^{-i\theta} &= \cos{\theta} - i\sin{\theta} \\ \\ e^{i\theta} - e^{-i\theta} &= 2i\sin{\theta} \\ \sin{\theta} &= \frac{e^{i\theta} - e^{-i\theta}}{2i} \\ \\ \sin{3x} &= \frac{e^{3i\,x} - e^{-3i\,x}}{2i} \\ \\ 3\sin{x} - 4\sin^3{x} &= 3\left(\frac{e^{i\,x} - e^{-i\,x}}{2i}\right) - 4\left(\frac{e^{i\,x} - e^{-i\,x}}{2i}\right)^3 \\ &= \frac{3e^{i\,x} - 3e^{-i\,x}}{2i} - 4\left(\frac{e^{3i\,x} - 3e^{i\,x} + 3e^{-i\,x} - e^{-3i\,x} }{-8i}\right) \\ &= \frac{3e^{i\,x} - 3e^{-i\,x}}{2i} + \frac{e^{3i\,x} - 3e^{i\,x} + 3e^{-i\,x} - e^{-3i\,x}}{2i} \\ &= \frac{e^{3i\,x} - e^{-3i\,x}}{2i} \\ &= \sin{3x} \end{align*}

    Q.E.D.
    Thank you very much do you by any chance know how to write ( -2 + i)e^(-pi/3) in the form a + bi?
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    Re: complex numbers eulers formula

    Quote Originally Posted by qwerty31 View Post
    Thank you very much do you by any chance know how to write ( -2 + i)e^(-pi/3) in the form a + bi?
    Yes, since \displaystyle \begin{align*} e^{\frac{-\pi}{3}} \end{align*} is a real number, just multiply each term in the brackets by it.
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    Re: complex numbers eulers formula

    Quote Originally Posted by Prove It View Post
    Yes, since \displaystyle \begin{align*} e^{\frac{-\pi}{3}} \end{align*} is a real number, just multiply each term in the brackets by it.
    Damn, sorry, i meant e^(-pi/3 i)
    I was thinking about turning the exponential into the form a + bi, then multiplying it by (-2 + i)?
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    Re: complex numbers eulers formula

    Quote Originally Posted by qwerty31 View Post
    Damn, sorry, i meant e^(-pi/3 i)
    I was thinking about turning the exponential into the form a + bi, then multiplying it by (-2 + i)?
    Yes, that's exactly what you would do.
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    Re: complex numbers eulers formula

    Quote Originally Posted by Prove It View Post
    Yes, that's exactly what you would do.
    thank you so much one more quick question, when solving square roots (or cube roots etc) of complex numbers, is there any specific patterns of what values of k to choose after equating the angles? I know for square roots the values of k taken are usually 0 and 1, but for cube roots some people take 0, 1, 2 and others take 0, 1, -1?
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    Re: complex numbers eulers formula

    Quote Originally Posted by qwerty31 View Post
    thank you so much one more quick question, when solving square roots (or cube roots etc) of complex numbers, is there any specific patterns of what values of k to choose after equating the angles? I know for square roots the values of k taken are usually 0 and 1, but for cube roots some people take 0, 1, 2 and others take 0, 1, -1?
    Whichever ones give the arguments as \displaystyle \begin{align*} \theta \in (-\pi, \pi] \end{align*}.
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