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Trigonometry question

  1. Nov 30, 2013 #1
    Hi, I have the following equation :

    2*sin3a=sqrt(2)

    I was able to solve it in a geometrical way(giving 6 solutions.) but I have no idea on how to solve it in a symbolic way...(which would take clearly less time to do ) Is there any way ?

    thank you
     
  2. jcsd
  3. Nov 30, 2013 #2

    SteamKing

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    If you know how to apply the arcsine, sure.
     
  4. Nov 30, 2013 #3
    Could you develop please ?I know arcsine... ( btw, it's from 0 to 2pi for the solutions)
     
  5. Nov 30, 2013 #4

    SteamKing

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    If sin(θ) = a, where -1 <= a <= 1; then

    arcsin(a) = θ, where θ is the principal angle
     
  6. Dec 1, 2013 #5

    haruspex

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    Do you happen to know an angle for which the sine is 1/√2?
     
  7. Dec 1, 2013 #6

    HallsofIvy

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    The point you seem to be missing is that if you divide both sides by 2 you have
    [tex]sin(3a)= \frac{\sqrt{2}}{2}= \frac{1}{\sqrt{2}}[/tex]
    (hence haruspex's question)

    Perhaps it would make more sense as "solve [itex]sin(\theta)= \sqrt{2}/2[/itex] for [itex]\theta[/itex]". Of course once you have found [itex]\theta[/itex], solve [itex]3a= \theta[/itex].

    (In order to find all a between 0 and [itex]2\pi[/itex], you may want to find all [itex]\theta= 3a[/itex] between 0 and [itex]6\pi[/itex].)
     
  8. Dec 1, 2013 #7

    CompuChip

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    There is a way that always works: if you can write the right hand side (##\tfrac{1}{\sqrt{2}}## in this case) as sin(b), then your equation is of the form

    sin(3a) = sin(b)

    and the general solution is

    $$3a = b + 2 \pi k, \text{ or } 3a = \pi - b + 2 \pi k$$

    for all ##k = 0, 1, -1, 2, -2, 3, -3, \ldots##.

    Similarly the general solution for cos(x) = cos(y) is

    $$x = y + 2 \pi k, \text{ or } x = - y + 2 \pi k$$

    (the only difference being the ##\pi## in the second branch).
     
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