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Trigonometric Equation

  1. Jan 15, 2005 #1
    Hello all

    Just refreshing in maths, and want to know if I am doing this correctly:

    Solve the equation [tex] \sin x = 0.2 [/tex] on the interval [tex] 0 \leq x < 2\pi [/tex].

    I took [tex] \arcsin(0.2) [/tex], however how do you solve for all solutions in the interval? (i.e what if the interval had been [tex] 0 \leq x < \pi [/tex])

    Thanks
     
  2. jcsd
  3. Jan 15, 2005 #2

    dextercioby

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    You know that "arcsin" only gives you the solution on the interval [itex] [-\frac{\pi}{2},\frac{\pi}{2}] [/itex],out of which you will only find useful the positive part.Okay,so that leaves you with another 3 intervals of width [itex] \frac{\pi}{2} [/itex].

    HINT:"arccos" will be useful for [itex] [\frac{\pi}{2},\pi] [/itex].
    Use that
    [tex] \cos(\frac{\pi}{2}-x) =\sin x [/tex]

    Daniel.

    PS.What about the interval [itex] (\pi,2\pi) [/itex] ??
     
  4. Jan 15, 2005 #3
    so [tex] \cos(\frac{\pi}{2}-x) =\sin x [/tex]

    then [tex] \arccos(\frac{\pi}{2}-x) = 0.2 [/tex] and then just solve for x.

    for [itex] [{\pi},[\frac{3\pi}{2}] [/itex]

    and [itex] [\frac{3\pi}{2}],[\frac{2\pi}] [/itex] would i employ the same reasoning?

    Thanks
     
  5. Jan 15, 2005 #4

    dextercioby

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    No,your eq.is wrong.
    [tex] \cos(\frac{\pi}{2}-x)=\sin x=0.2 [/tex]

    Solve for "x".

    Daniel.

    PS.On the second interval,the equation does not have solutions.Can u see why??
     
  6. Jan 15, 2005 #5
    hmmm...isn't there an easier way to do this? Shouldn't I just be able to use the period of [tex] \sin x [/tex] which is [tex] 2\pi [/tex] and divide it into 4 equal parts?

    Thanks
     
  7. Jan 15, 2005 #6

    dextercioby

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    Believe,me there ain't no simpler method.2 simple equations solvable via "arcsin" and "arccos" and the fact that,on the interval [itex] (\pi,2\pi) [/itex] the "sine" is negative will give you the answer.

    Daniel.
     
  8. Jan 15, 2005 #7

    Curious3141

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    Just use the fact that [tex]\sin x[/tex] is positive for [tex]x[/tex] in the 1st and 2nd quadrants. That means a domain of [tex] 0 \leq x < \pi [/tex]

    By plugging in [tex]\arcsin(0.2)[/tex], you'll be finding the 1st quadrant value. To get the other value in the second quadrant, just subtract the value of [tex]x[/tex] from [tex]\pi[/tex], et voila ! That is, if [tex]x[/tex] is one answer, then [tex](\pi - x)[/tex] is the other required answer.

    Another way of looking at this is :

    [tex]\sin (x) = \sin (\pi - x)[/tex]

    Equivalent expressions for cosine and tangent are :

    [tex]\cos(x) = \cos (2\pi - x) = \cos (-x)[/tex]

    [tex]\tan(x) = \tan (\pi + x)[/tex]
     
    Last edited: Jan 15, 2005
  9. Jan 15, 2005 #8

    dextercioby

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    Did u have a rough night???
    1.There's no such thing as [itex] \arcsin 2 [/itex]

    2.If he adds \pi to the first solution,he'll end up in the third quadrant...

    3.[tex] \sin(\pi+x)=-\sin x [/tex].

    If u don't know,at least don't show it...

    Daniel.

    EDIT:And one more thing:it's "VOILÀ",or "voilà"...
     
    Last edited: Jan 15, 2005
  10. Jan 15, 2005 #9

    Curious3141

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    I made a typo, genius. I just got up, Singapore time and everything.

    I know how to do this, you obviously don't. Honestly, WTF are you doing asking him to consider the arccosine in such a trivially simple problem ?!

    If YOU don't know, don't harangue others who do.
     
  11. Jan 15, 2005 #10

    dextercioby

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    You must have had a rough night... :tongue2: Was it because in Singapore the air it's more humid,hence it is harder to breathe?? :confused:

    You know them alright,you edited your post twice until got the final result.The last time was after i posted the reply in which i showed the mistakes...

    Daniel...

    PS.Expressions as "WTF" denote your level of intelligence... :yuck:
     
  12. Jan 15, 2005 #11

    Curious3141

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    OK, that's it ! :mad:

    FYI, I had just finished editing my post when you had waltzed in with you arrogant little "corrections". I don't need *you* to teach *me* trig.

    It's obvious that my knowledge of trig and possibly other areas of math exceeds yours. Might I remind you of that thread you started in the brainteaser section with a (to me) simple question, which I summarily solved, yet which you couldn't crack with your wonderful noggin ?

    I can see you are an arrogant, small minded phallus. You didn't bother to acknowledge my solution of your poser, because you're bearing a grudge after our run in about fundamental "axioms" in Physics. Which, BTW, I have yet to hear your learned dogma on.

    I can assure you my IQ is at least as high, if not higher than yours. Currently, I'm not even doing Maths/Physics/Engg as a primary discipline. My knowledge was gained many years back in school, with no practice in between. I think it's pretty good considering that.

    Keep your petty little taunts to the grade school yard where you belong.
     
  13. Jan 15, 2005 #12

    Integral

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    opps wrong button, The little red triange at the bottom left is the report message button.

    Both of your are getting a bit carried away. Just stop, I think the question has been answered.
     
    Last edited: Jan 15, 2005
  14. Jan 15, 2005 #13
    if we get the value of [tex] x [/tex] in the first two quadrants, that means there are no solutions in the third and fourth quadrants?

    Thanks
     
  15. Jan 15, 2005 #14

    Curious3141

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    Correct, there are no solutions for this equation in the 3rd and 4th quadrants. :smile:
     
  16. Jan 15, 2005 #15

    dextercioby

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    Yes,simply because the "sine" is negative,and your equation is asking for solutions which make "sine" 0.2,which is positive.

    Daniel.
     
  17. Jan 15, 2005 #16
    ok thanks a lot guys for taking your time to help me brush up on some trig!
     
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