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anonymous
 5 years ago
Can anybody point me to online resources that talk about using de moivre's theorem to prove trig identities like below:
sin 2x = 2 cos x. sin x
anonymous
 5 years ago
Can anybody point me to online resources that talk about using de moivre's theorem to prove trig identities like below: sin 2x = 2 cos x. sin x

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watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/users/watchmath#/users/watchmath/updates/4dd7c19dd95c8b0bdd3a61c4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm. the idea is this. \[e^{i\theta}=cos(\theta)+isin(\theta)\] and \[(e^i\theta)^2=e^{2i\theta}=(cos(\theta)+isin(\theta))^2=cos({2\theta})+isin(2\theta)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0square \[(cos(\theta) + i sin(\theta))\] to see what you get. then equate the real part to the real part and you get an identity for \[cos(2\theta)\] and another one for \[sin(2\theta)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://math.mit.edu/classes/18.013A/MathML/chapter18/section04.xhtml

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when you compute \[(cos(\theta)+isin(\theta))^2\] you get \[cos^2(\theta)sin^2(\theta)+i\times 2 cos(\theta)sin(\theta)\] the real part is \[cos^2(\theta)sin^2(\theta)\] so that must equal the real part of \[cos(2\theta)+isin(2\theta)\] which is just \[cos(2\theta)\] telling you that \[cosd(2\theta)=cos^2(\theta)sin^2(\theta)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0likewise \[sin(2\theta)=2cos(\theta)sin(\theta)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0My concern is: what de moivre's theorem says? z^n = r^n (( cos nx) + i sin (n x)) or (cos x + i sin x) ^n= ( cos nx) + i sin (n x) I am confused.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How come> can u explain?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[z=re^{i\theta}=r(cos(\theta)+isin(\theta))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[z^n=r^n(e^{i\theta})^n=r^ne^{ni\theta}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that by the laws of exponents.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and since \[e^{ni\theta}=cos(n\theta)+isin(n\theta) \] you get the second equality

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you have not seen \[z=re^{i\theta}\] as a representation of a complex number, then it requires a difffernt explanation, but if you have seen it it is nothing more than the laws of exponents.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0succinctly put here http://en.wikipedia.org/wiki/De_Moivre%27s_formula

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks .I need to site on this, it has been itching my head since yesterday.I appreciate your help.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0welcome hope at least second explanation was clear.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i must have made another algebra error let me check.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you have not seen \[z=re^{i\theta}\] as a representation of a complex number, then it requires a difffernt explanation, but if you have seen it it is nothing more than the laws of exponents.
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