# Teach me the best way of simplyfying this Long Demoivre's Trig Prooving question.

• Sep 24th 2008, 12:13 PM
afterahardtry
Teach me the best way of simplyfying this Long Demoivre's Trig Prooving question.
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.
• Sep 26th 2008, 06:46 AM
Opalg
For a start, the identities should be

$z-\frac1z = 2i\sin\theta$,

$z^n - z^{-n} = 2i\sin(n\theta)$.

You will also need the identity $z^n + z^{-n} = 2\cos(n\theta)$.

Then you can form the binomial expansion

\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.
• Sep 26th 2008, 07:09 AM
afterahardtry
Thank you so much (Rofl)
• Sep 27th 2008, 03:35 AM
bkarpuz
Quote:

Originally Posted by afterahardtry
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.

Let $z=r\mathrm{e}^{i\theta}$ for some $r\in[0,\infty)$ and $\theta\in[0,2\pi)$.
Then, we have $z=r(\cos(\theta)+i\sin(\theta))$.
For $n\in\mathbb{N}$, we clearly have
$[r(\cos(\theta)+i\sin(\theta))]^{n}=z^{n}$
............................._ $=[r\mathrm{e}^{i\theta}]^{n}$
............................._ $=r^{n}\mathrm{e}^{in\theta}$
............................._ $=r^{n}(\cos(n\theta)+i\sin(n\theta)).$
You may also need these.
• Sep 28th 2008, 11:52 PM
afterahardtry
Quote:

Originally Posted by Opalg

Then you can form the binomial expansion

\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.

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 .
• Sep 29th 2008, 12:06 AM
Opalg
Quote:

Originally Posted by afterahardtry
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 .

I don't know whether this is the cause of the dificulty, but I wonder if you are overlooking the "i" in $(2i\sin\theta)^6$. Since $i^6=-1$, this expression becomes $-64\sin^6\theta$.