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How to solve trig equations?

  1. Jul 14, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]8cos2\theta-4\sqrt{3}=0[/tex]


    2. Relevant equations

    I don't have one.

    3. The attempt at a solution

    I am really not sure where to begin, our lesson tonight was lacking detail.

    I guess start by addding [tex]4\sqrt{3}[/tex] to the right side

    Then what? I don't think I should divide by [tex]8[/tex] because that represents [tex]8\ of\ cos2\theta[/tex] and that's where I'm lost???
     
  2. jcsd
  3. Jul 14, 2011 #2

    SammyS

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    Yes, you divide (both sides) by 8, because that represents 1/8 of 8 times [itex]\cos(2\theta)[/itex], which is [itex]\cos(2\theta)[/itex].
     
  4. Jul 14, 2011 #3
    Thanks, I was just trying it that way.

    I'm starting to get it.
     
  5. Jul 17, 2011 #4
    How do you get the actual notation(square roots and equals sign)...do you have an application where you can write mathematical notation, greek alphabet etc???
     
  6. Jul 17, 2011 #5
    If you chose to reply using the "go advanced" option when adding a reply or when you create a new topic click on the sigma button. This button is the first row right above the field in which you enter text and closest to your right. From there it should be pretty self explanatory. I don't use it often. I find it easier to just use plain text.
     
    Last edited: Jul 17, 2011
  7. Jul 17, 2011 #6
    I used tex, this was actually my first time using it.

    https://www.physicsforums.com/showthread.php?t=8997

    and thanks for the help with the problem, I am still struggling with last part of similar problems, like when to add [tex]2\pi(n)[/tex] versus adding [tex]pi(n)[/tex].
     
  8. Jul 17, 2011 #7
    Tangent is a periodic function that's periodic with respect to pi. Cosine and sine are periodic functions with respect to 2*pi.
     
  9. Jul 17, 2011 #8
    I understand that. I think the wording of our homework, and going back and forth from radian to degree just had me a little lost. I think I just need to slow down. I missed some due to small errors.

    Take this problem for example: [tex]sin(t+\pi)=sint[/tex]
    I picked the answer that said: Not an identity because it is false at [itex]t=\pi[/itex]
    but the correct answer was: At t=0 the equation is true.

    If the equation is true anywhere, then should that be my answer?
    I was thinking, if it was false anywhere, then that was my answer?

    I was torn between these 2 answers, and did not know which one trumped the other. Ah the joys of an online class.
     
  10. Jul 17, 2011 #9
    sin(t+pi)=sin(t)
    use the fact that sin(t+pi)=-sin(t)
    -sin(t)=sin(t)
    add sin(t) to both sides
    0=2*sin(t)
    divide both sides by 2
    0=sin(t)
    sin^-1 both sides
    0 = t
    lastly
    0+2*pi*n=t
    were
    n is the set of integers
    or simply
    2*pi*n=t
    and if you let n=0 then you get 0=t

    so the equation is true at all the values that occur at t=2*pi*n were n is the set of integers (this includes negatives integers, zero, and positive integers) and false on t values not equal to this...

    as far as the answer choices... I don't think I understand the English lol... it is indeed false at certain t values and true on certain t values so i guess if the answer choices were

    is true anywhere
    is false anywhere

    then both are technically correct??? =_=
     
  11. Jul 17, 2011 #10
    After reviewing the problem several times, I have found my mistake, I missed the small print that stated: Select a true statement about the equation.
     
  12. Jul 17, 2011 #11
    Actually, the equation is true at [tex]t=\pi[/tex]
    as
    [tex]\sin(2\pi) = 0 = \sin(\pi)[/tex]
    In fact, the equation is true at: [tex]t=n\pi[/tex] Can you see why?
     
  13. Jul 17, 2011 #12
    [tex]sin(t+\pi)=sint[/tex][tex]sin(\pi+\pi)=sin\pi[/tex][tex]sin2\pi=sin\pi[/tex]
    [itex]1 \not= -1[/itex] is not true
     
    Last edited: Jul 17, 2011
  14. Jul 17, 2011 #13
    hm what i do wrong? lets see...
     
  15. Jul 17, 2011 #14
    Oh I see, I used cos instead of sin. It should be 0=0.
     
  16. Jul 17, 2011 #15
    It's not that you're doing anything wrong, just that you've omitted a step. Remember that [tex]\sin(\pi-x)=\sin(x)[/tex] Graphically, this means that any y coordinate on the unit circle (with the exception of y = -1 or 1) exists on either side of the y-axis. Your equation solved for the y coordinates to the right of the y-axis. Can you see how to solve for those on the left?
     
  17. Jul 17, 2011 #16

    Ray Vickson

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    Just as a matter of interest: when using "tex" there is a difference between "sin" and "\sin" (same for "cos" vs. "\cos", "log" vs. "\log", etc.). Here they are, for comparison: [itex] sin(\theta)[/itex] vs. [itex]\sin(\theta)[/itex].

    RGV
     
  18. Jul 17, 2011 #17
    Not really...

    The sin is the y value, so the sin of say [itex]\frac{5\pi}{2}[/itex] is 1/2, but when I add pi it becomes -1/2

    The original equation was [itex]\sin(t+\pi)=\sin(t)[/itex] not [tex]\sin(\pi-t)[/tex]
     
  19. Jul 17, 2011 #18
    Thanks, I am just learning this.
     
  20. Jul 17, 2011 #19
    Oh, sorry, I introduced [itex]\sin(\pi-x)=\sin(x)[/itex] as a completely new equation, which says that the unit circle has reflectional symmetry across the y-axis. It's useful for solving trig equations because it shows how the solutions relate. For example, if [itex]\sin(x)=k[/itex] has a solution in Quadrant I, it must also have a solution in Quadrant II, because we can reflect Quadrant I over the y-axis to get Quadrant II.

    On the other hand, [itex]\sin(\pi+t)=\sin(t)[/itex] is only true for some values of t. Graphically, it can be interpreted as saying "The y coordinate of some point at angle t along the unit circle is the same as the y coordinate of some other point, located a half-turn away from t (because pi radians is 180 degrees, or half of a full revolution)." Can you see why this means only multiples of [itex]\pi[/itex] are solutions?
     
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