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Proving De Moivre's (1 Viewer)

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yeah you just treat sin3x and cos3x as sin(2x+x) and cos(2x+x) and use your rules to expand until you get to an expression in terms of sinx and cosx. It's a bit long, but not too hard if you know your trigonometric properties
 

jet

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Well, De Moivre's theorem doesn't actually take into account the modulus.

The best way would be to show
[r(cosx + isinx)]2 = r2(cos(2x) +isin(2x))

And then multiply it by r(cosx + isinx) again. If they wanted you to actually 'prove' De Moivre's by induction, they'd ask for it.
 

addikaye03

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But you couldn't use the complex numbers method for finding the triple angle results in THIS question. That would make your reasoning circular. It would involve using exactly the result you are trying to prove.
Well i mean't it's the fastest way to find the sin3x and cos3x results. It wouldn't be considered as necessary working, I think if you said: "Consider cos3x= whatever it is", then that would be fine. I don't think it would be considered as part of the proof, it's a pretty well known result.
 
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jet

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We can't actually use Euler's formula, which is the stupidest thing ever.
 
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But you don't have to be stupid just because your teacher is stupid.
 
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There's no need to prove Euler's formula every time you use it.

Do you prove the fundamental theorem of calculus every time you calculate an integral?
 
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If you don't know the fundamental theorem of calculus, perhaps I should put it another way.

Do you prove Pythagoras' Theorem every time you use it?
 

jet

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Fundamental theorem of calculus has 2 parts:

I) Differentiation and integration are the reverse of each other.

II) The definite integral is equal to the difference between the primitives on the bounds of the integral.

That's the gist of it, without being too rigourous. A bitch to prove as well :p
 

cutemouse

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But Jetblack I recall your teacher took marks off for something like not putting +/- for finding a polynomial equation with roots a^2, b^2 and c^2 (where a, b and c are roots of the original polynomial equation).

That's also stupid IMO.
 

jet

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Yes she did. it was fair, I wasn't rigourous enough. Of course I was disappointed, but I never forgot that mistake again :p
 

cutemouse

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Oh come on.... :p It's a method commonly employed and it never matters if u take negative or positive since you end up squaring the expression to make it a polynomial.
 

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