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($\cos \phi + i\sin \phi)^n = \cos (n \phi) + i\sin (n \phi)$

Saw this in the De Moivre's formula proof and some other calculations involving complex numbers, but I do not understand why the equation is true.

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    Did you mean $(\cos \phi + i\sin \phi)^n = \cos (n \phi) + i\sin (n \phi)$? – 2017-01-15
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    Yes, sorry I haven't considered it's specific to DeMoivre's formula. Edited the question, I saw it in some calculations involving complex numbers but still can't understand why it stands. – 2017-01-15

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The "identity is wrong. Consider $n=2$. Then the left hand side is $$ \cos^2 \theta + \sin^2 \theta + 2\cos \theta \sin\theta = 1 + 2\cos \theta \sin\theta = 1 + \sin(2\theta) $$ And the right hand side is $$ \cos (2\theta) + \sin(2\theta) $$ So unless $\cos (2\theta) = 1$ the equality fails.

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    You did not take the complex unit into account. – 2018-05-29
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    @MrYouMath When I posted my answer, the complex $i$ was not present in the problem. An hour later, at 6:18, an edit was made to the problem inserting the $i$. So you have downvoted me for what was a correct answer to the problem as posed at the time of the answer. – 2018-05-31
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    Sorry, I just looked at the time of answer and compared it with the answer of Simply Beautiful Art did not compare it with the edited time. I cannot remove the downvote at the moment because you would need to edit your answer. I know it is frustrating when the OPs change the question. Still, it is not useful to let this answer here when it is not adapted to the edited question. – 2018-05-31
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This is DeMoivre's formula:

$$(\cos\phi+i\sin\phi)^n=\cos n\phi+i\sin n\phi$$

which may be proven by induction:

First of all, this is true for $n=1$. Now, we prove that if it is true for $k$, then it is true for $k+1$:

$$\begin{align}(\cos\phi+i\sin\phi)^{k+1}&=(\cos\phi+i\sin\phi)(\cos\phi+i\sin\phi)^k\\&=(\cos\phi+i\sin\phi)(\cos k\phi+i\sin k\phi)\\&=\cos\phi\cos k\phi-\sin\phi\sin k\phi+i(\sin\phi\cos k\phi+\cos\phi\sin k\phi)\\&\stackrel{(*)}=\cos(k+1)\phi+i\sin(k+1)\phi\end{align}$$

That is, if it holds for $k=1$, it holds true for $k+1=2$, and if it holds for $k=2$, it holds true for $k+1=3$, etc.

$(*)$ we used the sum of angles formula.

Likewise, this extends to negative $n$:

$$\begin{align}(\cos\phi+i\sin\phi)^{-n}&=((\cos\phi+i\sin\phi)^n)^{-1}\\&=(\cos n\phi+i\sin n\phi)^{-1}\\&=\frac1{\cos n\phi+i\sin n\phi}\\&=\frac1{\cos n\phi+i\sin n\phi}\frac{\cos n\phi-i\sin n\phi}{\cos n\phi-i\sin n\phi}\\&=\frac{\cos n\phi-i\sin n\phi}{\cos^2n\phi+\sin^2n\phi}\\&\stackrel{(**)}=\cos n\phi-i\sin n\phi\\&\stackrel{(***)}=\cos(-n\phi)+i\sin(-n\phi)\end{align}$$

$(**)$ we used the pythagorean identity $\sin^2+\cos^2=1$

$(***)$ we used symmetry formulas $\cos(x)=\cos(-x)$ and $\sin(x)=-\sin(-x)$

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    Yes, edited the question that's the line I don't get. I understand all the other parts of the induction, but this step is still not clear. – 2017-01-15
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    @Jani I've updated on the induction. – 2017-01-15
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DeMoivre's formula.

$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$

De Moivre's formula can easily be derived from Euler's formula.

$ e^{i \theta} =(\cos \theta + i \sin \theta)$

$ e^{i n \theta} =(\cos n \theta + i \sin n \theta)$

And you can use induction to prove it. Here's link with full details.

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    Haha, and Euler's formula can easily be derived from...wait, that's not easier to prove! – 2017-01-15
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    Thanks, I finally put it together, so the step in the DeMoivre proof works because of Euler's formula which has its own separate proof. – 2017-01-15
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    @Simple Art I think you need to read full article from Wikipedia. – 2017-01-15
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    Your welcome... – 2017-01-15
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    @KanwaljitSingh What do you mean? I can clearly understand everything there, and I wasn't saying this is wrong. I was merely pointing out that proving something like Euler's formula is not elementary. – 2017-01-15
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    As someone once said, when I ask for why $$1782^{12}+1841^{12}\ne1922^{12}$$I hope the answer is the left side is odd and the right side is even and not "because it contradicts Fermat's last theorem". If you chose the second path, I expect you to prove Fermat's last theorem for me. – 2017-01-15
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    @Jani But proof of Euler's formula is nowhere near elementary my dear fellow. – 2017-01-15
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    @Simple Art this question is not about to prove Euler's formula. – 2017-01-15
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    You are just guiding him to wrong side by including Euler proof. – 2017-01-15
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    That's right, this question is not about proving Euler's formula, and anyone who understands the proof of Euler's formula likely understands the proof of DeMoivre's formula, but anyone who doesn't understand the proof of DeMoivre's formula likely doesn't even understand Euler's formula. To prove Euler's formula, you should start somewhere along the lines of what properties analytic functions have, and that is well out of scope of this question. – 2017-01-15
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    @Simple Art I accepted this answer because the only problematic line for me in the DeMoivre proof was the one in question, based on Euler's formula. ( I understand it's not trivial to prove but for my purposes I can just accept it as true ). My basic misunderstanding was I thought it might be some trigonometric identity which turned out to be not the case. – 2017-01-16
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    @Simple Art also I appreciate you typing out the first half of the DeMoivre proof but that was actually the clear part, since for n>0 integers it doesn't require Euler's formula. It only gets tricky once we try prove it for n<0 as well. – 2017-01-16
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    @Jani Hm, ok. I added that part into my answer for negative values. And I think when I say this, I speak for both Kanwaljit and I: "Euler's formula is awesome, so please learn it anyways :D" – 2017-01-16
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    Agree with you bro :-) – 2017-01-16