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Trig Form Conversion question

  1. May 20, 2008 #1
    1. The problem statement, all variables and given/known data

    This is a solutuion to a differential equation I've solved:

    [tex] -cos(\sqrt(2)t) - 2sin(\sqrt(2)t) [/tex]

    I've been asked to put it in the form:

    [tex] Asin(wt + phi) [/tex]


    2. Relevant equations

    [tex] A= sqrt(-1^2 + -2^2) [/tex]

    [tex] phi = sin^{-1}phi = \frac{c1}{A} [/tex]
    [tex] phi = cos^{-1} phi = \frac{c2}{A} [/tex]
    and
    [tex] phi = tan^{-1} phi = \frac{c1}{c2} [/tex]


    3. The attempt at a solution


    A = 5

    [tex] phi = sin^{-1}phi = \frac{-1}{\sqrt(5)} [/tex]
    [tex] phi = -0.4636[/tex]

    [tex] phi = cos^{-1} phi = \frac{-2}{\sqrt(5)} [/tex]
    [tex] phi = 2.678[/tex]

    and
    [tex] phi = tan^{-1} phi = \frac{-1}{-2} [/tex]
    [tex] phi = 0.4636[/tex]

    Can you help with the phase angle? (3 different answers - none of which are correct)

    The book has 3.6052 which I see is pi + 0.4636.

    Thanks
    -Sparky
     
  2. jcsd
  3. May 21, 2008 #2

    Defennder

    User Avatar
    Homework Helper

    A isn't 5. You miswrote your answer here, despite having found the correct one.

    The "basic angle" (do you use that term? I don't know if this is the proper term) of [tex]\phi[/tex] is 0.4636.

    [tex]sin \ \phi = -\frac{1}{\sqrt{5}}[/tex] Are you familiar with the trigonometric quadrants ASTC, where A is the top right hand quadrant and the rest are labeled in that order anti-clockwise?

    From that diagram, sin is negative for the bottom 2 quadrants. That means to say the angle phi is either [tex]\pi + \phi \ \mbox{or} \ -\phi[/tex]'

    [tex]cos \ \phi = -\frac{2}{\sqrt{5}} [/tex] Again you want the values of the angle for which cos phi is negative. From the ASTC quadrant diagram, cosine is negative for [tex]\pi - \phi \ \mbox{or} \ \pi + \phi[/tex]

    The only possible solution which satisfies both is [tex]\pi + \phi[/tex].
     
  4. May 21, 2008 #3
    Oops, typing error, I meant [tex] A= \sqrt(-1^2 + -2^2) [/tex]

    [tex] A= \sqrt(5) [/tex]

    Thanks
     
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