Hi there,i've attached the question in the image...It's basically some test question that takes my time, just being curious can one simplify my technique? thank you very much.
Hi there,i've attached the question in the image...It's basically some test question that takes my time, just being curious can one simplify my technique? thank you very much.
For a start, the identities should be
$\displaystyle z-\frac1z = 2i\sin\theta$,
$\displaystyle z^n - z^{-n} = 2i\sin(n\theta)$.
You will also need the identity $\displaystyle z^n + z^{-n} = 2\cos(n\theta)$.
Then you can form the binomial expansion
$\displaystyle \begin{aligned}(2i\sin\theta)^6 = (z-z^{-1})^6 &= z^6 -6z^4 + 15z^2 - 20 + 15z^{-2} - 6z^{-4} + z^{-6}\\ &= (z^6+ z^{-6}) - 6(z^4 + z^{-4}) + 15(z^2 + z^{-2}) - 20\\ &= 2\cos(6\theta) - 12\cos(4\theta) + 30\cos(2\theta) -20,\end{aligned}$
and the result follows easily.
Let $\displaystyle z=r\mathrm{e}^{i\theta}$ for some $\displaystyle r\in[0,\infty)$ and $\displaystyle \theta\in[0,2\pi)$.
Then, we have $\displaystyle z=r(\cos(\theta)+i\sin(\theta))$.
For $\displaystyle n\in\mathbb{N}$, we clearly have
$\displaystyle [r(\cos(\theta)+i\sin(\theta))]^{n}=z^{n}$
............................._$\displaystyle =[r\mathrm{e}^{i\theta}]^{n}$
............................._$\displaystyle =r^{n}\mathrm{e}^{in\theta}$
............................._$\displaystyle =r^{n}(\cos(n\theta)+i\sin(n\theta)).$
You may also need these.
Dear sir, for some reasons i would have a negative integer ,example
1/32[ xxxx - 10] ,
in oppose of the proove where it is seeking for
1/32[ xxxx + 10] ,
from what i have learn -sin (theta) = - sin (theta) , and -cos (theta) = cos ( theta) which i could not apply in this theorem, is there a way for changing the the proove without jeopardizing sin^6 into negative? thank you very much for the help .