1. ## Demonstration envolving Complex numbers!! Help please

Please someone help me.. I've an exam in 2 days ... The following problem makes me nuts.. I can't get it right..

Show that:

cos(∏ - alpha) + i cos (∏/2 - alpha)
-------------------------------------------- = cis (∏ - 2alpha)
cis (alpha)

I'm desperate... I've tried everything I know.. any help is appreciated

2. ## Re: Demonstration envolving Complex numbers!! Help please

Originally Posted by josepbigorra
Please someone help me.. I've an exam in 2 days ... The following problem makes me nuts.. I can't get it right..

Show that:

cos(∏ - alpha) + i cos (∏/2 - alpha)
-------------------------------------------- = cis (∏ - 2alpha)
cis (alpha)

I'm desperate... I've tried everything I know.. any help is appreciated
What have you tried? It's hard to give you advice when we don't know what you've tried.

I'd favor expanding out $\displaystyle cos( \pi - \alpha ) = cos(\pi)~cos(\alpha) + sin(\pi)~sin(\alpha) = -cos(\alpha)$ etc.

-Dan

3. ## Re: Demonstration envolving Complex numbers!! Help please

What I've tried is the following:
$cos (\pi- \alpha ) = - cos (\alpha )$
$i cos (\frac{\pi }{2} - \alpha ) = isin (\alpha )$
Then i proceed to the substitution in the expression from above, but I don't know how to continue.

4. ## Re: Demonstration envolving Complex numbers!! Help please

Originally Posted by topsquark
What have you tried? It's hard to give you advice when we don't know what you've tried.

I'd favor expanding out $\displaystyle cos( \pi - \alpha ) = cos(\pi)~cos(\alpha) + sin(\pi)~sin(\alpha) = -cos(\alpha)$ etc.

-Dan
What I've tried is the following:

Then i proceed to the substitution in the expression from above, but I don't know how to continue.

5. ## Re: Demonstration envolving Complex numbers!! Help please

Okay, you've got the $\displaystyle cos( \pi - \alpha )$ and $\displaystyle sin( \pi / 2 - \alpha )$ expressions. So you need to evaluate:
$\displaystyle \frac{-cos( \alpha ) + i ~sin( \alpha ) }{cos( \alpha ) + i~sin(\alpha)}$

You could do the division but recall Euler's theorem: $\displaystyle cos(x) + i~sin(x) = e^{ix}$. See what you can do with this.

-Dan

6. ## Re: Demonstration envolving Complex numbers!! Help please

Or "rationalize the denominator" by multiplying numerator and denominator by the conjugate of the denominator:
$\displaystyle \left(\frac{-cos(\alpha)+ isin(\alpha)}{cos(\alpha)+ isin(\alpha)}\right)\left(\frac{cos(\alpha)- isin(\alpha)}{cos(\alpha)- isin(\alpha)}\right)$
$\displaystyle = -cos^2(\alpha)+ sin^2(\alpha)+ 2sin(\alpha)cos(\alpha)$.

Now, what about that left side? $\displaystyle cis(\pi- 2\alpha)= cos(\pi- 2\alpha)+ isin(\pi- 2\alpha)= -cos(2\alpha)+ isin(2\alpha)$
So what are $\displaystyle cos(2\alpha)$ and $\displaystyle sin(2\alpha)$?

7. ## Re: Demonstration envolving Complex numbers!! Help please

Originally Posted by HallsofIvy
Or "rationalize the denominator" by multiplying numerator and denominator by the conjugate of the denominator:
$\displaystyle \left(\frac{-cos(\alpha)+ isin(\alpha)}{cos(\alpha)+ isin(\alpha)}\right)\left(\frac{cos(\alpha)- isin(\alpha)}{cos(\alpha)- isin(\alpha)}\right)$
$\displaystyle = -cos^2(\alpha)+ sin^2(\alpha)+ 2sin(\alpha)cos(\alpha)$.

Now, what about that left side? $\displaystyle cis(\pi- 2\alpha)= cos(\pi- 2\alpha)+ isin(\pi- 2\alpha)= -cos(2\alpha)+ isin(2\alpha)$
So what are $\displaystyle cos(2\alpha)$ and $\displaystyle sin(2\alpha)$?

My friend thanks a lot for the info... I managed to get the problem working.. I really appreciate your help