Using eulers formula show that sin3x = 3sinx - 4sin^3x
How do I do that?
No need for Euler's formula.
You know that $\displaystyle \displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^n = \sum_{r = 0}^n {n\choose{r}}\cos^{n-r}{\theta}\sin^r{\theta} \end{align*}$ by the Binomial Theorem, and $\displaystyle \displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^n = \cos{n\theta} + i\sin{n\theta} \end{align*}$ by DeMoivre's Theorem.
Try applying each of these to $\displaystyle \displaystyle \begin{align*} \left(\cos{\theta} + i\sin{\theta}\right)^3 \end{align*}$, setting them equal, and equating real and imaginary parts
Hmm, well let's see...
$\displaystyle \displaystyle \begin{align*} e^{i\theta} &= \cos{\theta} + i\sin{\theta} \\ \\ e^{-i\theta} &= \cos{\theta} - i\sin{\theta} \\ \\ e^{i\theta} - e^{-i\theta} &= 2i\sin{\theta} \\ \sin{\theta} &= \frac{e^{i\theta} - e^{-i\theta}}{2i} \\ \\ \sin{3x} &= \frac{e^{3i\,x} - e^{-3i\,x}}{2i} \\ \\ 3\sin{x} - 4\sin^3{x} &= 3\left(\frac{e^{i\,x} - e^{-i\,x}}{2i}\right) - 4\left(\frac{e^{i\,x} - e^{-i\,x}}{2i}\right)^3 \\ &= \frac{3e^{i\,x} - 3e^{-i\,x}}{2i} - 4\left(\frac{e^{3i\,x} - 3e^{i\,x} + 3e^{-i\,x} - e^{-3i\,x} }{-8i}\right) \\ &= \frac{3e^{i\,x} - 3e^{-i\,x}}{2i} + \frac{e^{3i\,x} - 3e^{i\,x} + 3e^{-i\,x} - e^{-3i\,x}}{2i} \\ &= \frac{e^{3i\,x} - e^{-3i\,x}}{2i} \\ &= \sin{3x} \end{align*}$
Q.E.D.
thank you 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?