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First step is

$$\cos5\theta + i \sin 5\theta = (\cos \theta + i \sin \theta)^5$$

thanks

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  • $\begingroup$ Because it's easier than using using other trigonometric formulae. Note that the real parts of the expressions on each side of the equation must be equal. $\endgroup$ – user164587 Sep 26 '14 at 14:39
  • $\begingroup$ @user164587 well i meant how to get it? $\endgroup$ – problematic Sep 26 '14 at 14:41
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    $\begingroup$ It is de Moivre $\endgroup$ – Seub Sep 26 '14 at 14:41
  • $\begingroup$ @vera sorry bad english $\endgroup$ – problematic Sep 26 '14 at 14:41
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Hint: Expand $(\cos\theta + i\sin\theta)^5$ using binomial theorem. i.e. $(a+b)^n = \sum\limits_{k = 0}^n {n \choose k}a^k b^{n-k}$
Here, $a = \cos\theta, b=i\sin\theta$.

$$\begin{align}(\cos\theta+i\sin\theta)^5 &= \sum\limits_{k = 0}^5 {5\choose k}(\cos\theta)^k (i\sin\theta)^{5-k}\\ &= {5\choose0}(\cos\theta)^0(i\sin\theta)^5 + {5\choose1}(\cos\theta)^1(i\sin\theta)^4 + {5\choose2}(\cos\theta)^2(i\sin\theta)^3\\&\quad+ {5\choose3}(\cos\theta)^3(i\sin\theta)^2+{5\choose4}(\cos\theta)^4(i\sin\theta)^1+{5\choose5}(\cos\theta)^5(i\sin\theta)^0 \\&= (i\sin\theta)^5 + 5(\cos\theta)(i^4\sin^4\theta) + 10\cos^2\theta(i^3\sin^3\theta)\\&\quad+10\cos^3\theta(i^2\sin^2\theta)+5\cos^4\theta(i\sin\theta)+\cos^5\theta \\&=\cdots \end{align}$$

Then, for even powers of $\sin \theta$, use $\sin^2 \theta = 1-\cos^2 \theta$.

Finally, equate the real parts of $\cos(5\theta)+i\sin(5\theta) = (\cos\theta+i\sin\theta)^5$.

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  • $\begingroup$ am i doing it right ? lol $(_0^5)$$(cos^0 \theta)$$(i sin^5 \theta)$ $\endgroup$ – problematic Sep 26 '14 at 15:03
  • $\begingroup$ Yes, and there will be $5$ more terms. $\endgroup$ – taninamdar Sep 26 '14 at 15:05
  • $\begingroup$ actually i don't know how binomial theoram works, can you give an example and show me what $(_0^5)$$(cos^0 \theta)$$(sin^5 \theta)$ goes to, i will do the rest my self $\endgroup$ – problematic Sep 26 '14 at 15:06
  • $\begingroup$ See the edit, ask if you need more help. $\endgroup$ – taninamdar Sep 26 '14 at 15:13
  • $\begingroup$ thank you, how to get the first number? i mean it looks like $(_1^5)$ : 5x1, $(_2^5)$ 5x2 , then $(_3^5)$ onward? $\endgroup$ – problematic Sep 26 '14 at 15:22
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The point is that $$\cos 5 \theta$$ is the even part of $$\exp(5 i \theta) = [\exp(i \theta)]^5 = (\cos \theta + i \sin \theta)^5,$$ where the second equality follows from de Moivre's Theorem. Then, by expanding the right-hand side using the Binomial Theorem, we can immediately write $\cos 5 \theta$ as a sum of products of powers of $\sin \theta$ and $\cos \theta$. Then, we can eliminate even powers of $\sin \theta$ using the Pythagorean identity $$\sin^2 \theta + \cos^2 \theta = 1.$$

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  • $\begingroup$ You're welcome, I hope you found it helpful. $\endgroup$ – Travis Willse Sep 26 '14 at 15:37
  • $\begingroup$ yes, you showed me exp(5iθ)=[exp(iθ)]5=(cosθ+isinθ)5, i needed this to know why we have the first step, i can't choose 2 accepted answers... $\endgroup$ – problematic Sep 26 '14 at 15:40
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    $\begingroup$ No problem, that's just a part of the site mechanics, the important thing is that you understand the concepts you asked about better than you did before. $\endgroup$ – Travis Willse Sep 26 '14 at 15:51
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De Moivre's Theorem states $$\cos n\theta + \mathbb i \sin n\theta = \left( \cos \theta + \mathbb i \sin \theta \right)^n$$

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