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Math Help - Trig

  1. #1
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    Trig

    tana^2 = sin(2a)

    tried everything... -_-
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  2. #2
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    Quote Originally Posted by ruthvik View Post
    tana^2 = sin(2a)

    tried everything... -_-

    Do you mean:

     tan(a^2) = sin(2a)

    or

     tan^2(a) = sin(2a)

    ?????

    Doesn't really matter, neither of them are true.
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  3. #3
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    Quote Originally Posted by Mush View Post
    Do you mean:

     tan(a^2) = sin(2a)

    or

     tan^2(a) = sin(2a)

    ?????

    Doesn't really matter, neither of them are true.
    this happened in different topic
    im trying to solve for solutions
    not proving it
    and i meant this (tana) ^ 2 = sin(2a)
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  4. #4
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    \tan^2 a = \sin (2a)

    We'll make use of the identity: \sin (2a) = \frac{2\tan a}{1 + \tan^2 a}

    So:
    \begin{aligned} \tan^2 a & = \frac{2\tan a}{1 + \tan^2 a} \\ \tan^2 a (1 + \tan^2 a) & = 2\tan a \\ & \ \vdots \\ \tan^4 a + \tan^2 a - 2\tan a & = 0 \\ \tan a \left(\tan^3 a + \tan a - 2\right) & = 0 \\ \tan a (\tan a - 1) (\tan^2 a + \tan a + 2) & = 0 \end{aligned}

    Set each factor equal to 0 and solve for a. Note that the third factor (essentially a quadratic) does not have real solutions for \tan a so don't worry about it.
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  5. #5
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    Quote Originally Posted by o_O View Post
    \tan^2 a = \sin (2a)

    We'll make use of the identity: \sin (2a) = \frac{2\tan a}{1 + \tan^2 a}

    So:
    \begin{aligned} \tan^2 a & = \frac{2\tan a}{1 + \tan^2 a} \\ \tan^2 a (1 + \tan^2 a) & = 2\tan a \\ & \ \vdots \\ \tan^4 a + \tan^2 a - 2\tan a & = 0 \\ \tan a \left(\tan^3 a + \tan a - 2\right) & = 0 \\ \tan a (\tan a - 1) (\tan^2 a + \tan a + 2) & = 0 \end{aligned}

    Set each factor equal to 0 and solve for a. Note that the third factor (essentially a quadratic) does not have real solutions for \tan a so don't worry about it.
    this was perfect but the thing was this is bonus
    which means he probably expects to us to do out whole thing
    should i just do the last one too? by using quadratic forumula?
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  6. #6
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    Yes . Or simply look at the discriminant and it'll show you that there are no real solutions to the quadratic.
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  7. #7
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    Quote Originally Posted by o_O View Post
    Yes . Or simply look at the discriminant and it'll show you that there are no real solutions to the quadratic.
    i dont know how to do trig with imaginary numbers
    i dont think he will except no real solutions for that one
    ill try
    thanks a lot
    btw is there any way to prove sin2a equals the thing u gave me?
    aight i got tana = o, pi, pi/4, 5pi/4
    Last edited by ruthvik; December 17th 2008 at 04:01 PM.
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  8. #8
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    Exactly. That's why there are no solutions that are real numbers. There is no angle such that when you plug it into \tan^2 a + \tan a + 2 you will get 0.

    To prove the identity: \sin (2a) = \frac{2\tan a}{1+\tan^2a}

    Why don't you try ? It isn't too difficult really. Start off with using: \sec^2 x = 1 + \tan^2 x
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  9. #9
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    Quote Originally Posted by o_O View Post
    Exactly. That's why there are no solutions that are real numbers. There is no angle such that when you plug it into \tan^2 a + \tan a + 2 you will get 0.

    To prove the identity: \sin (2a) = \frac{2\tan a}{1+\tan^2a}

    Why don't you try ? It isn't too difficult really. Start off with using: \sec^2 x = 1 + \tan^2 x
    nah i just said i changed the right side to tan by multiplying top and bottom by cosa^2

    so 0,pi/4, pi, 5pi/4 r my solutions since
    tana = 0 , a can only equal = {0, pi} in the interval i was given
    tana = 1, a can only equal = (5pi/4, pi/4} in the interval between 0 and 2pi noninclusive
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