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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

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  1. watchmath
    • 5 years ago
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    http://openstudy.com/users/watchmath#/users/watchmath/updates/4dd7c19dd95c8b0bdd3a61c4

  2. anonymous
    • 5 years ago
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    hmm. the idea is this. \[e^{i\theta}=cos(\theta)+isin(\theta)\] and \[(e^i\theta)^2=e^{2i\theta}=(cos(\theta)+isin(\theta))^2=cos({2\theta})+isin(2\theta)\]

  3. anonymous
    • 5 years ago
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    square \[(cos(\theta) + i sin(\theta))\] 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 \[sin(2\theta)\]

  4. anonymous
    • 5 years ago
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    http://math.mit.edu/classes/18.013A/MathML/chapter18/section04.xhtml

  5. anonymous
    • 5 years ago
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    when 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 \[cosd(2\theta)=cos^2(\theta)-sin^2(\theta)\]

  6. anonymous
    • 5 years ago
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    likewise \[sin(2\theta)=2cos(\theta)sin(\theta)\]

  7. anonymous
    • 5 years ago
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    My 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.

  8. anonymous
    • 5 years ago
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    both are true.

  9. anonymous
    • 5 years ago
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    How come> can u explain?

  10. anonymous
    • 5 years ago
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    \[z=re^{i\theta}=r(cos(\theta)+isin(\theta))\]

  11. anonymous
    • 5 years ago
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    \[z^n=r^n(e^{i\theta})^n=r^ne^{ni\theta}\]

  12. anonymous
    • 5 years ago
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    that by the laws of exponents.

  13. anonymous
    • 5 years ago
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    and since \[e^{ni\theta}=cos(n\theta)+isin(n\theta) \] you get the second equality

  14. anonymous
    • 5 years ago
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    if 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.

  15. anonymous
    • 5 years ago
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    succinctly put here http://en.wikipedia.org/wiki/De_Moivre%27s_formula

  16. anonymous
    • 5 years ago
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    Thanks .I need to site on this, it has been itching my head since yesterday.I appreciate your help.

  17. anonymous
    • 5 years ago
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    welcome hope at least second explanation was clear.

  18. anonymous
    • 5 years ago
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    i must have made another algebra error let me check.

  19. anonymous
    • 5 years ago
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    if 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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