Complex Numbers! (1 Viewer)

harshsyd · Jul 26, 2018

Hey, I got two questions;

I'll attach the images as I cannot use LaTeX for some reason.

1. https://imgur.com/a/CFbQErW (I tried expanding and simplifying using bionomial, I do get correct answers but of course they are not in the forms of cot and so on, I get straight square roots, expected).

2. https://imgur.com/a/eF2Su32 (The only real place I am struggling is the last part, deducing the cos(2O) statement, rest is completed).

Thank You!

Drongoski · Jul 26, 2018

harshsyd · Jul 26, 2018

Thank you buddy, that was very nicely explained!
Except, how was the second last line derived? I understood up till 1+cosO+isinO/1-cosO-isinO...the half angles one, I am slightly lost

fan96 · Jul 26, 2018

$1 + \cos \theta = \left(\sin^2{\theta\over2} + \cos^2{\theta\over2}\right)+\left( \cos^2{\theta\over2} - \sin^2{\theta\over2}\right) = 2 \cos^2{\theta\over2}$

$1 - \cos \theta = \left(\sin^2{\theta\over2} + \cos^2{\theta\over2}\right)-\left( \cos^2{\theta\over2} - \sin^2{\theta\over2}\right) = 2 \sin^2{\theta\over2}$

I think the start of that second last line you were talking about should read:

$\frac{2 \cos^2 {\theta \over 2}+ i(2 \sin{\theta \over 2} \cos {\theta \over 2})}{2 \sin^2 {\theta \over 2} - i(2\sin{\theta \over 2}\cos{\theta \over 2})}$

fan96 · Jul 26, 2018

For your second question:

Let $z = {\rm cis\,} \theta$ .

${\rm cis \,} 4\theta + 1 = ({\rm cis\,} 2\theta - \sqrt2{\rm cis\,}\theta + 1)({\rm cis\,} 2\theta + \sqrt2{\rm cis\,}\theta + 1)$

Note:
$\begin{aligned} ({\rm cis\,} 2\theta + \sqrt2{\rm cis\,}\theta + 1) &=[(\cos 2\theta + 1 + \sqrt 2 \cos \theta) + i(\sin 2\theta + \sqrt 2 \sin\theta) \\ &= (2 \cos^2 \theta + \sqrt 2 \cos \theta) + i(2\sin \theta \cos\theta+ \sqrt 2 \sin\theta)\\ &= 2\cos\theta( \cos \theta + {\sqrt2/2}) + i[2 \sin \theta(\cos\theta + {\sqrt2/2})]\end{aligned}$

Similarly,
$({\rm cis\,} 2\theta - \sqrt2{\rm cis\,}\theta + 1) = 2\cos\theta( \cos \theta- {\sqrt2/2}) + i[2 \sin \theta(\cos\theta - {\sqrt2/2})]$

Going back to the original identity: multiplying the expression on the RHS and taking the real part of both sides,

$\begin{aligned}\cos4\theta + 1 &= 4 \cos^2\theta( \cos^2 \theta - (\sqrt2/2)^2) - 4 \sin^2\theta( \cos^2 \theta - (\sqrt2/2)^2) \\ &= 4 \cos2\theta( \cos \theta + {\sqrt2/2}) ( \cos \theta - {\sqrt2/2}) \end{aligned}$

And a few more lines of working give the required identity.

harshsyd · Jul 27, 2018

Thanks!

Complex Numbers! (1 Viewer)

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