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Math Help - Trigo Equations/Identities Problems

  1. #1
    gp3
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    Trigo Equations/Identities Problems

    Hi guys,

    I've been studying the notes that my teacher gave me but i still can't solve these few questions!!!

    Can someone help me? I have a test coming up next week! Thanks a lot.



    Ans:


    Ans:
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  2. #2
    Super Member bigwave's Avatar
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    Talking it becomes a quadratic

    cot^2\theta + csc\theta = 1
    use the identity cot^2\theta = csc^2\theta - 1
    so,
    csc^2\theta - 1 + csc\theta = 1
    set it to zero
    csc^2\theta + csc\theta -2 = 0
    this factors
    (csc\theta - 1)(csc\theta +2) = 0
    from this you will get
    \frac{\pi}{2}, \frac{11\pi}{6}, \frac{7\pi}{6}

    do you understand the answers from the quadradric
    Last edited by bigwave; November 29th 2009 at 10:56 PM. Reason: latex, wording
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  3. #3
    gp3
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    now one thing I don't really understand is why is it that some of these questions only have 1 answers while some have more than 5!

    I know you have to look at which quadrant they are in and the range, is that true? for eg. in the 2nd quad, you have to take 180deg + 'basic angle'. I'm not really clear on this.
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  4. #4
    Super Member bigwave's Avatar
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    ok starting from the factored quadradic which is
    (csc\theta - 1)(csc\theta +2) = 0
    if you take (csc\theta - 1) so that it becomes zero the whole equation goes to zero so if csc\theta = 1 then that will be some our answer

    csc\theta = \frac{1}{sin\theta}

    so \frac{1}{sin\theta} = 1 or  sin\theta = 1 so sin\theta is only 1 at \frac{\pi}{2} that is only place on the 2\pi interval that it can

    we do the same with (csc^2\theta + 2) except the results will show up in 2 places on the 2\pi interval

    can you see this....
    sorry i am slow on this not real fast with the latex
    I will help you more if you need.
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  5. #5
    gp3
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    sorry man, i dont really understand: 'only place on the interval'

    and means the first quadrant?

    how did you factor to get ?

    I use the quadratic formula and only get .


    really thanks for your help!

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  6. #6
    Super Member bigwave's Avatar
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    2\pi interval is 0 \le \theta \le2\pi

    \frac{2}{\pi} is within 0 \le \theta \le2\pi we are not concerned about what quadrant it is in.

    csc^2\theta + csc\theta -2 = 0
    using
    u substitution where u = csc\theta
    you have

    u^2 + u - 2 = 0

    which is factored as
    (u - 1)(u +2)

    it is easier to use sin\theta rather than csc\theta since we can then recognise the commonly known \theta
    so csc\theta = \frac{1}{sin} so substituting it back into the factors we can solve
    see the chart below(which i stole from another thread to see how these were derived..

    are we getting the picture???


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  7. #7
    gp3
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    Smile

    thanks buddy! I'll try to see if i can solve it first.
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  8. #8
    gp3
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    alright i got the first ans.

    but how do you get the other 2 answers since gives you a negative answer (-2), and not in the range, I thought we must leave it out, as in we cant get an answer from it?
    Last edited by gp3; November 30th 2009 at 05:15 AM.
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  9. #9
    Super Member bigwave's Avatar
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    Quote Originally Posted by gp3 View Post
    alright i got the first ans.

    but how do you get the other 2 answers since gives you a negative answer (-2), and not in the range, I thought we must leave it out, as in we cant get an answer from it?
    ok, again, it will be easier to replace
    csc^2\theta with the identity \frac{1}{sin^2\theta}

    then we can work with \frac{1}{sin^2\theta} = -2

    multiply both sides by sin^2\theta to get 1 = -2sin\theta

    now divide both sides by -2 to get -\frac{1}{2} = sin\theta
    this has 2 answers \frac{7\pi}{6}and \frac{11\pi}{6}
    they are in the III and IV quadrants respectfully

    do you still need help with the next one you originally posted
    Last edited by bigwave; November 30th 2009 at 05:06 PM. Reason: latex
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  10. #10
    gp3
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    hey really big thanks man! I understand how to solve it now!

    and i got how to solve the other one as well. Thanks!
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