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For the function $$G(w) = \frac{\sqrt2}{2}-\frac{\sqrt2}{2}e^{iw},$$ show that $$G(w) = -\sqrt2ie^{iw/2} \sin(w/2).$$ Ive been told to use the equation below by use of the complex sine definition, but i would like to know the exact steps to get there and then how to simplify it to get my result.

$-\sqrt{2}ie^{i\frac{w}{2}}\sin\frac{w}{2} =-\sqrt{2}ie^{i\frac{w}{2}}(\frac{1}{2i}(e^{i\frac{w}{2}} - e^{-i\frac{w}{2}})) $

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Really? Just multiply it out. I provided the answer a while ago... –  Euler....IS_ALIVE Aug 25 '12 at 5:17
    
This is your third question about this problem (math.stackexchange.com/questions/186123/… and math.stackexchange.com/questions/184491/…). Much better to have just edited the first one than to have asked two more. Keep all the info together in one place and let people see what's already been done. –  Gerry Myerson Aug 25 '12 at 5:34

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$$\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}e^{iw} = \frac{\sqrt{2}}{2}(1 - e^{iw})$$

then pull out $e^{\frac{iw}{2}}$ and you get

$$ = \frac{\sqrt{2}}{2}e^{\frac{iw}{2}}\left(e^{-\frac{iw}{2}} - e^{\frac{iw}{2}}\right) $$

then pull out one $(-i)$ and push the $\frac{1}{2}$ inside. "Pull out one $(-i)$" might seem slightly weird since there was no real factor of $-1$ or $i$ multiplied in the entire expression, so what were really doing is multiplying by $1$ in a very convenient way. The reasoning behind multiplying by $1$ in this way is that we know the formula for complex sin ($\sin(x) = \frac{e^{ix} -e^{-ix}}{2i}$) and this clever multiplication allows us to get it.

So then we have

$$ = -\sqrt{2}ie^{\frac{iw}{2}}\left(\frac{e^{\frac{iw}{2}} - e^{-\frac{iw}{2}}}{2i}\right) $$

so now since in our last step we set up our equation to have the formula for complex sine, we'll now substitute the complex sine and we have

$$= -\sqrt{2}ie^{\frac{iw}{2}}\sin\left(\frac{w}{2}\right) $$

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