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Math Help - Trig Equation

  1. #1
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    Trig Equation

    Hello everyone, this is my first post here and I am very glad to have discovered this forum. I will try very hard to get the Latex correct, but please forgive me if I mess something up.

    I am very stuck on the following question.

    <br />
3sin(2\theta)=cos(2\theta)<br />

    This is how I have tried to solve it using Trig identities.

    <br />
3(2sin\theta cos\theta)=sin^2\theta-cos^2\theta<br />
    <br />
6sin\theta cos\theta-(sin^2\theta+cos^2\theta)=0<br />
    <br />
sin\theta(6cos\theta)-1=0<br />
    I have a hunch that I should just stop here, because I can tell I've done something wrong. I would continue on trying to find sin=1 and cos=1, but it just doesn't work.

    Any help that can be offered would be greatly appreciated. Also, \\ isn't giving me a new line in latex. Perhaps I am missing something there?

    Thanks,
    Sleight
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  2. #2
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    My mistake, I believe that sin^2\theta-cos^2\theta should be cos^2\theta-sin^2\theta.

    It is still getting me into the same place though.
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  3. #3
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    Assuming that values of \displaystyle \theta that make \displaystyle \cos{2\theta} = 0 are not possible solutions (I'll leave you to check this)

    \displaystyle \frac{3\sin{2\theta}}{\cos{2\theta}} = 1

    \displaystyle 3\tan{2\theta} = 1

    \displaystyle \tan{2\theta} = \frac{1}{3}.

    Go from here...
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  4. #4
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    reply

    hello dear
    try to do it like this:
    suppose tanx=3
    find x by using calculator
     tanx=sinx/cosx
    multiply the whole equation by sinx
    can u take it from here..
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  5. #5
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    Wow, thank you both for the very quick answer. Looking at that now it seems obvious, however I have one small hang-up in the directions of my textbook, and I am not quite sure what it means. It reads like this.
    "In algebraic equations, we generally do not divide each side by any expression that involves a variable, and the same rule holds for trigonometric equations."
    Because of that sentence (whatever that means) I haven't been cross dividing anything.

    So if I follow the workings below,

    <br />
3tan2\theta=1<br />

    <br />
tan2\theta=\frac{1}{3}<br />

    <br />
2\theta = 18.43...<br />

    <br />
\theta = 9.22...<br />

    Strange though, if that is the correct track. Every other answer has been giving me two answers. I suppose if I shifted that into quadrant 3 I would get 12.36.

    Still on the right track?
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  6. #6
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    There will be more than one answer...

    Assuming you have used radians...

    Since the tangent function is positive in the first and third quadrants, if \displaystyle \tan{2\theta} = \frac{1}{3} then

    \displaystyle 2\theta = \left\{\arctan{\frac{1}{3}}, \pi + \arctan{\frac{1}{3}}\right\} + 2\pi n, where \displaystyle n is an integer.
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  7. #7
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    Yes, I have used radians. One part of this problem that I forgot to mention is that it is in the interval [0,2╥), so I don't need +2k╥. Sorry, I should have mentioned that earlier.

    I also noticed that I made an error in my last post. It should look like this.

    <br />
tan2\theta= \frac{1}{3}<br />

    <br />
2\theta = tan^-^1(\frac{1}{3}) = \approx.322<br />

    <br />
\theta = \approx0.16<br />

    This works when I plug it back in, so thank you very much. I was not able to get this far on my own. Taking it further, I suppose that I can find the other again within the interval by looking at the unit circle. Doing this I find that the other angle is approximately 3.30..

    You guys rock.

    Any advice on the // thing for spacing in Latex? I am doing it in sort of a long-about fashion. Which you can see if you look at my code.
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  8. #8
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    You actually do need the \displaystyle + 2\pi n, because if

    \displaystyle 2\theta = \left\{\arctan{\frac{1}{3}}, \pi + \arctan{\frac{1}{3}}\right\} + 2\pi n then

    \displaystyle \theta = \left\{\frac{1}{2}\arctan{\frac{1}{3}}, \frac{\pi}{2} + \frac{1}{2}\arctan{\frac{1}{3}}\right\} + \pi n.

    It's quite possible that since the period is diminished, then there are more solutions in the interval.
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  9. #9
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    Your completely right, I had forgotten about the half angle identity involved. Although it appears that these answers are only in quadrant 1 and 3, so they are already ╥ distance away from each other.

    Once again, I appreciate all of your help.
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