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Math Help - evaluating trig ratios

  1. #1
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    evaluating trig ratios

    hello, i am trying to evaluate a trig ratio but the example is quite hard to follow. can someone please explain how this question is done,


    given that sinx=\frac{12}{13}, and that \frac{\pi}{2}< x<\pi
    show that  cos x= \frac{-5}{13}
    hence without calculators evaluate sin (x +\pi) and cos2x
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  2. #2
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    Quote Originally Posted by sigma1 View Post
    hello, i am trying to evaluate a trig ratio but the example is quite hard to follow. can someone please explain how this question is done,


    given that sinx=\frac{12}{13}, and that \frac{\pi}{2}< x<\pi
    show that  cos x= \frac{-5}{13}
    hence without calculators evaluate sin (x +\pi) and cos2x
    Dear sigma1,

    For the first part use, cosx=\pm\sqrt{1-sin^{2}x} but since theta is in the second quadrant the cosine must be negative hence you can leave out the positive sign.

    Hope this will help you.
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  3. #3
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    Quote Originally Posted by Sudharaka View Post
    Dear sigma1,

    For the first part use, cosx=\pm\sqrt{1-sin^{2}x} but since theta is in the second quadrant the cosine must be negative hence you can leave out the positive sign.

    Hope this will help you.

    thanks i now see the identity butcan you show me how it is applied to prove the question and later evaluate the given trig ratios. thanks.. didnt understand it in class.
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  4. #4
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    Quote Originally Posted by sigma1 View Post
    thanks i now see the identity butcan you show me how it is applied to prove the question and later evaluate the given trig ratios. thanks.. didnt understand it in class.
    Dear sigma1,

    All you got here is to substitute the values. I think you can do that. For the second part, what you should know is that sin(x+\pi)=-sinx and cos2x=cos^{2}x-sin^{2}x

    A more detailed discription about trignometric identities could be found at, List of trigonometric identities - Wikipedia, the free encyclopedia

    Hope this helps.
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  5. #5
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    ok from what i can understand i

    substituted  <br />
\sqrt{1-sin^2x} = \frac{-5}{3}<br /> <br />
     <br />
1-sin^2x =\frac {-5}{13}^2<br />

    simplifying that should prove the answer right. but i cant seem to simplify it to prove the answer. i must be doing something wrong

    because i get sinx= \frac{8}{13}
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  6. #6
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    Quote Originally Posted by sigma1 View Post
    hello, i am trying to evaluate a trig ratio but the example is quite hard to follow. can someone please explain how this question is done,


    given that sinx=\frac{12}{13}, and that \frac{\pi}{2}< x<\pi
    show that  cos x= \frac{-5}{13}
    hence without calculators evaluate sin (x +\pi) and cos2x
    hi

    Since x falls in the second quadrant , sin is positive and cos is negative

    Draw a triangle with sides labelled according to the trigo ratio , and use phythagoras theorem to evaluate the other side .

    in this case , hypothenus=13 , opposite side =12, adjacent=5

    cos x=adjacent/hypothenus=5/13 but look at red , so cos x=-5/13

    for the remaining examples ,just expand and substitute appropriately .
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  7. #7
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    thanks alot i evaluated the others easily from that...
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