Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 5"
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\left(\sqrt{2}\,\text{cis}\,\frac{\pi}{4}\right)^{17} - \left(\sqrt{2}\,\text{cis}\,-\frac{\pi}{4}\right)^{17} &= 2^{17/2}\,\left(\text{cis}\,\frac{17\pi}{4}\right) - 2^{17/2}\,\left(\text{cis}\,-\frac{17\pi}{4}\right) \\ | \left(\sqrt{2}\,\text{cis}\,\frac{\pi}{4}\right)^{17} - \left(\sqrt{2}\,\text{cis}\,-\frac{\pi}{4}\right)^{17} &= 2^{17/2}\,\left(\text{cis}\,\frac{17\pi}{4}\right) - 2^{17/2}\,\left(\text{cis}\,-\frac{17\pi}{4}\right) \\ | ||
| − | &= 2^{17/2}\left | + | &= 2^{17/2}\left[\text{cis}\left(\frac{\pi}{4}\right) - \text{cis}\left(-\frac{\pi}{4}\right)\right] \\ |
| − | &= 2^{17/2}\left( | + | &= 2^{17/2}\left(2i\sin \frac{\pi}{4}\right) \\ |
| − | &= 2^{17/2} \cdot 2 \cdot 2^{-1/2} = 2^ | + | &= 2^{17/2} \cdot 2 \cdot 2^{-1/2}i = 2^9i = \boxed{512}\,i |
\end{align*}</cmath> | \end{align*}</cmath> | ||
, where 
. Find 
.
and 
, where 
. By ![\begin{align*} \left(\sqrt{2}\,\text{cis}\,\frac{\pi}{4}\right)^{17} - \left(\sqrt{2}\,\text{cis}\,-\frac{\pi}{4}\right)^{17} &= 2^{17/2}\,\left(\text{cis}\,\frac{17\pi}{4}\right) - 2^{17/2}\,\left(\text{cis}\,-\frac{17\pi}{4}\right) \\ &= 2^{17/2}\left[\text{cis}\left(\frac{\pi}{4}\right) - \text{cis}\left(-\frac{\pi}{4}\right)\right] \\ &= 2^{17/2}\left(2i\sin \frac{\pi}{4}\right) \\ &= 2^{17/2} \cdot 2 \cdot 2^{-1/2}i = 2^9i = \boxed{512}\,i \end{align*}](http://latex.artofproblemsolving.com/7/9/8/79867d703166170eb8ad773a0383b65941927cd7.png)
