De Moivre question (complex numbers)

TsAmE · **Joined:** Mon, 19 Apr 2010 13:45:05 UTC **Posts:** 61

De Moivre's Theorem says that if $\theta$ is real, then $(cos\theta + isin\theta)^n = cos(n\theta) + isin(n\theta)$ for all n E Z.

By considering real parts, find an expression for $cos(5\theta)$ in terms of powers of $cos\theta$ and $sin\theta$ .

Attempt:

$cos(5\theta) = cos^2(2.5\theta) - sin^2(2.5\theta)$

but this wasnt the correct answer.

outermeasure · **Posted:** Sat, 23 Oct 2010 15:17:13 UTC

TsAmE wrote:

De Moivre's Theorem says that if $\theta$ is real, then $(cos\theta + isin\theta)^n = cos(n\theta) + isin(n\theta)$ for all n E Z.

By considering real parts, find an expression for $cos(5\theta)$ in terms of powers of $cos\theta$ and $sin\theta$ .

Attempt:

$cos(5\theta) = cos^2(2.5\theta) - sin^2(2.5\theta)$

but this wasnt the correct answer.

Read the question again. What can you tell me about $(\cos\theta+i\sin\theta)^5$ ?

peccavi_2006 · **Posted:** Sun, 24 Oct 2010 01:29:44 UTC

Okay, I want to have a stab at this one:

$\\ \cos (5 \theta) + i \sin(5 \theta) = (\cos(\theta) + i \sin(\theta))^{5} \\ = \cos ^{5} (\theta) + 5 i \cos ^{4} (\theta) \sin (\theta) + 10 i^{2} \cos ^{3} (\theta) \sin ^{2} (\theta) + 10 i^{3} \cos^{2} (\theta) \sin^{3} (\theta) + 5 i^{4} \cos(\theta) \sin^{4} (\theta) + i^{5} \sin ^{5} (\theta) \\ = \cos^{5}(\theta) + 5i \cos^{4}(\theta) \sin(\theta) - 10 \cos^{3}(\theta) \sin^{2}(\theta) - 10i \cos^{2}(\theta) \sin^{3}(\theta) + 5 \cos(\theta) \sin^{4}(\theta) + i \sin^{5}(\theta)$

Now we want the real parts, which is everything without an i in it. So
$\cos(5 \theta) = \cos (\theta) ( \cos ^{4} (\theta) - 10 \cos^{2}(\theta) \sin^{2}(\theta) + 5 \sin^{4}(\theta) )$

Then, ofc, as $\sin^{2}(\theta) = 1 - \cos^{2}(\theta)$ you could write it as a sum of cosines, but if your answer is in sines and cosines then the above should be fine

TsAmE · **Joined:** Mon, 19 Apr 2010 13:45:05 UTC **Posts:** 61

Thanks a lot. For the last question it asked write $sin(5\theta)$ in terms of powers of $sin\theta$ .

In this case how would this be possible since $sin(5\theta)$ is imaginary?

peccavi_2006 · **Posted:** Sun, 24 Oct 2010 14:24:55 UTC

For $\sin(5 \theta)$ take all of the imaginary terms in the expansion of $(\cos(\theta) + i \sin(\theta))^{5}$ . Then use $\cos^{2}(\theta) = 1 - \sin^{2}(\theta)$ to rewrite the terms with cosines in them

Shadow · **Posted:** Sun, 24 Oct 2010 15:55:38 UTC

TsAmE wrote:

Thanks a lot. For the last question it asked write $sin(5\theta)$ in terms of powers of $sin\theta$ .

In this case how would this be possible since $sin(5\theta)$ is imaginary?

It's not imaginary, it's the imaginary part of the complex number. If you don't get it you should remember that x+iy is complex, but the point is that y itself is real.

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De Moivre question (complex numbers)

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