Solve de Moivre's Theorem: Sin 3x & Cos 3x

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The discussion focuses on using de Moivre's Theorem to derive expressions for sin 3x and cos 3x, leading to the identity for tan 3x. Participants confirm that tan 3x can be set to 1 to solve the cubic equation t^3 - 3t^2 - 3t + 1 = 0. They identify that the roots of the equation correspond to values of x where tan(3x) equals 1, specifically finding x values of π/12, 5π/12, and 9π/12. The conversation concludes with the realization that t = 1 is not a root, and one of the roots is t = -1, confirming the correct solutions.
  • #31
Well... I think so...
 
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  • #32
sooyong94 said:
Well... I think so...

Can't think why not. You've found three roots with the trig identity that you correctly got through deMoivre. They all work. Cubic equations have a maximum of three roots. We've disposed of the idea t=1 is one of them. Can't think of what else could go wrong.
 
  • #33
Thanks then. :D
 
  • #34
sooyong94 said:
Thanks then. :D

Very welcome.
 

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