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I'm an idiot in trig

  1. Oct 23, 2005 #1
    ok i have no idea how to do this, and i'm pretty sure I should.
    i have to express this in terms of the general sine funtion

    f(x) = -2sin3x - 4cos3x

    I don't even know where to start
     
  2. jcsd
  3. Oct 23, 2005 #2
    Expressions of that format [tex]a\sin x + b\cos x[/tex] can be rewritten in terms of a sine only for positive values of a and b.

    However, you have negative values for a and b, so i'd start by multiplying throughout by -1.

    Then you could say:

    [tex]a\sin x + b\cos x \equiv R\sin(x - \alpha)[/tex]

    Expand the right hand side of the equation and equate coefficients to find R and alpha.
     
    Last edited: Oct 23, 2005
  4. Oct 23, 2005 #3

    Pyrrhus

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    Work out a right triangle with -2 and -4, where the hypothenuse is [itex] A = \sqrt{(-2)^2 + (-4)^2} [/itex] directed an angle [itex] \phi [/itex].
     
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