# Thread: Complex Algebra (De Moivre?)

1. ## Complex Algebra (De Moivre?)

Hello,

Would you use De Moivres theorem here and expand it? Not to sure!

Thanks.

By writing sin theta in terms of z = e ^(i theta), express sin^6 theta as a sum of
terms of the form k cos(h theta)

2. You don't need to write it in terms of $\displaystyle \displaystyle e^{i\theta}$.

If $\displaystyle \displaystyle z = \cos{\theta} + i\sin{\theta}$, then

$\displaystyle \displaystyle z^6 = (\cos{\theta} + i\sin{\theta})^6$.

Expand this using the Binomial Theorem and write in terms of real and imaginary parts.

You should also know that $\displaystyle \displaystyle z^6 = \cos{6\theta} + i\sin{6\theta}$ by DeMoivre's Theorem.

So once you have written the binomial expansion in terms of its real and imaginary parts, you can then equate real and imaginary parts and solve the problem.

3. Originally Posted by MattWT
Hello,

Would you use De Moivres theorem here and expand it? Not to sure!

Thanks.
By writing sin theta in terms of z = e ^(i theta), express sin^6 theta as a sum of
terms of the form k cos(h theta)
I would write $\displaystyle \sin\theta = \frac1{2i}(z-z^{-1})$, where $\displaystyle z=e^{i\theta}$. Then $\displaystyle \sin^6\theta = \bigl(\frac1{2i}\bigr)^6(z-z^{-1})^6$. Expand that by the binomial theorem, then pair off terms from opposite ends of the expansion and use facts like $\displaystyle z^6+z^{-6} = 2\cos6\theta$ (by DeMoivre's Theorem).