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Math Help - pre-calc/trig

  1. #1
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    pre-calc/trig

    My teacher always confuses me when she throws T into the mix. i tried replacing with X, didnt help my problem either. i saw in a thread by yovivo, that the help posted could somewhat be applied. so i tried for the first one how the identities might help. kinda went off and got lost again... so im just exhausted with these four problems. thanks for any help..


    Directions: Find all values for t in the interval [0,2PI]
    1. cot t = csc t
    2. sin^2 t - cos^2 t= 1
    3.|csct|=2
    4. sin^2 t =3/4
    Juan
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  2. #2
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    1. \cot{t} = \csc{t}

    \Rightarrow \frac{\cos{t}}{\sin{t}} = \frac{1}{\sin{t}}

    \Rightarrow \cos{t} \sin{t} =\sin{t}

    \Rightarrow \sin{t}( \cos{t} - 1) = 0

    Find values of t for \sin{t} = 0 and \cos{t} = 1 in the interval [0,~2\pi] for your solutions.


    2. \sin^2{t} - \cos^2{t} = \cos{(2t)}<br />
    So solve \cos{(2t)} = 1.


    3. absolute of \csc{t} = 2

    \Rightarrow absolute of \sin{t} = \frac{1}<br />
{2}

    Solve \sin{t} = \frac{1}{2} \mbox{and} \sin{t} = -\frac{1}{2}


    4. If sin^2{t} = \frac{3}{4}

    \Rightarrow \sin{t} = \frac{\sqrt{3}}{2}<br />
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  3. #3
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    ok all seems good but.....
    for the fist one, i cant see how you go from trig identities to costsint=sint. i tried multiplying but on the left i get 0. so from that point on im lost..

    then on the second one, is it as simple as divide by 2?

    also for these solutions, do i out pi at the end? the first one is confusing me, do i just say that t is 0 and 2pi? but aren't they the same?

    then is it truly like 1/2? so 1/2pi for 3?

    Juan
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  4. #4
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    I've just realised that it should read:

    \cot{t} = \csc{t}

    \Rightarrow \frac{\cos{t}}{\sin{t}} = \frac{1}{\sin{t}}

    \Rightarrow \cos{t} =1

    \Rightarrow t = 0,~2\pi

    No they are not the same, but there are 2 different solutions to the equation.

    For #2,

    if \cos{(2t)} = 1, then 2t = 0 \mbox{ or } 2\pi

    For #3, solving \sin{t} = \frac{1}{2} \mbox{ we get } t = \frac{\pi}{6},~ \frac{5\pi}{6}

    Solving \sin{t} = -\frac{1}{2} \mbox{ we get } t = \frac{7\pi}{6},~\frac{11\pi}{6}


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  5. #5
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    so cos(2t) = cos (t) ??
    maybe im just over thinking it?
    Juan
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  6. #6
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    If you solve \cos{(2t)} = 1, you get 2t = 0 or 2t = 2\pi or 2t = 4\pi, which means your solutions are t = 0 and t = \pi and t = 2\pi

    Look at a cosine graph and you will see that at 0,~2\pi,~4\pi,.... \quad \cos{t} = 1

    The graph of \cos{(2t)} is 2 cosine cycles squashed into the interval [0,~2\pi]. So 2t = 0,~2\pi,~4\pi,.... which means t = 0,~\pi,~2\pi,....
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  7. #7
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    thanks! i see what you are talking about now!
    actually my teacher never showed us a sine cycle and all that good stuff....

    Juan
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  8. #8
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    Slight typo here:
    2.
    It should say \sin^2t-\cos^2t = -\cos(2t)<br />
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