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)
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.
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).