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Trigonometry - sinusoid

  1. Jul 8, 2016 #1

    Rectifier

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    The problem

    I am trying to write ##-sin 2x + \sqrt{3} \cos 2x ## as ## A \sin(2x+\phi)## without using this formula below ##a sin (x) + b cos (x) = c sin (x+v), \ \ c=\sqrt{a^2+b^2} \ ,\ \ \tan v = \frac{b}{a}##

    The attempt

    ##A \sin(2x+\phi) = A \cos(\phi) \sin(2x) + A \sin(\phi) \cos(2x) ## comparison with the starting expression.

    [tex]\begin{cases} A \cos(\phi) = -1 \\ A \sin(\phi) = \sqrt{3} \end{cases}[/tex]

    ## \phi : ##
    [tex]\frac{A \sin(\phi)}{A \cos(\phi)} = \tan \phi = \frac{\sqrt{3}}{-1} = - \sqrt{3} \\ \tan -\phi = \sqrt{3} \\ \phi = -\frac{\pi}{3} + \pi n, \ \ n \in \mathbb{Z}[/tex]

    For ## n=0 ## -> ##-\frac{\pi}{3}## gives $$A \sin(-\frac{\pi}{3}) = \sqrt{3} \\ -A \sin(\frac{\pi}{3}) = \sqrt{3} \\ -A \frac{\sqrt{3}}{2}=\sqrt{3} \\ A = -2$$

    ##n = 1## from above gives ## \phi = \frac{2 \pi}{3} ##
    $$ A \sin(\phi) = \sqrt{3} \\ A \sin(\frac{2 \pi}{3}) = \sqrt{3}$$

    I can rewrite ## \sin(\frac{2 \pi}{3}) ## as ## 2 \sin(\frac{\pi}{3}) \cos(\frac{\pi}{3}) = 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}##

    $$ A \sin(\frac{2 \pi}{3}) = \sqrt{3} \\ A \frac{\sqrt{3}}{2} = \sqrt{3} \\ A = 2$$

    Which A should I use in my answer? And why?
     
  2. jcsd
  3. Jul 8, 2016 #2
    An easier approach
    $$-sin 2x + \sqrt{3}cos 2x = 2\left ( -\frac{1}{2}sin 2x+\frac{\sqrt{3}}{2} cos 2x\right )=2\left ( cos (2\pi/3) sin 2x+sin(2\pi/3)cos 2x\right )=2sin(2x + 2 \pi/3)$$
     
  4. Jul 8, 2016 #3

    Rectifier

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    Thank you for that elegant solution. I solved the problem myself just now, it does not really matter which A I use as long as I use it with the corresponding angle, since ## -2 \sin(2x- \frac{\pi}{3}) = 2 \sin(2x + \frac{2 \pi}{3}) ## .
     
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