Mod-arg to x+iy

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Mod-arg to x+iy

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RobertDyd

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Mod-arg to x+iy

« on: March 04, 2017, 10:17:39 am »

Hi,

I was wondering if there was a way of converting "sqrt(2)[Cos(-17pi/12) +iSin(-17pi/12)]" into x+iy form without getting decimals, the question I am doing wants the answer to be in exact form.

Thanks,
YourLocalRob

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RuiAce

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Re: Mod-arg to x+iy

« Reply #1 on: March 04, 2017, 10:27:19 am »

$\text{The process is not immediate}$
$\text{To find the exact ratios for }\frac{5\pi}{12}\\ \begin{align*}\cos \frac{5\pi}{12}&=\cos \left(\frac{\pi}{6}+\frac{\pi}{4}\right)\\ &= \cos \frac{\pi}{6}\cos \frac{\pi}{4}-\sin \frac{\pi}{6}\sin \frac{\pi}{4}\\ &= \frac{\sqrt3-1}{2\sqrt{2}}\\ \sin \frac{5\pi}{12}&=\sin \frac{\pi}{6}\cos \frac{\pi}{4}+\cos \frac{\pi}{6}\sin \frac{\pi}{4}\\ &= \frac{\sqrt3+1}{2\sqrt2}\end{align*}$
$\text{Then, since }-\frac{17\pi}{12}\text{ is a negative 3rd quadrant angle}\\ \text{(equivalently 2nd quadrant angle)}\\ \text{with related angle }\frac{5\pi}{12}\\ \sqrt{2}\left( \cos \left(-\frac{17\pi}{12}\right) + i \sin \left(-\frac{17\pi}{12}\right)\right)=\sqrt{2}\left(-\frac{\sqrt3-1}{2\sqrt{2}}+i \frac{\sqrt3+1}{2\sqrt2}\right)$

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Author Topic: Mod-arg to x+iy (Read 1349 times) Tweet Share

RobertDyd

Mod-arg to x+iy

RuiAce

Re: Mod-arg to x+iy

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