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Math Help - taking trig class and forgot how to do this problem.

  1. #1
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    taking trig class and forgot how to do this problem.

    haven't taken math in a while and I'm taking a trigonometry class. I believe i learned this in algebra but i can't remember how its done.

    cos = +/-√1-225/289

    (basically i don't understand how to do the square root part of this. its the square root of 1- 225 over 289.

    The answer for this is = +/- 8/17

    **no clue how they got 8/17.
    thx and godbless if anybody can help me

    http://i48.photobucket.com/albums/f2...799/sqroot.jpg
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  2. #2
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    Think about sinē(θ) + cosē(θ) = 1 and tell us what you get from that.
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  3. #3
    Super Member bigwave's Avatar
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    add together

    Quote Originally Posted by coryjannis View Post
    haven't taken math in a while and I'm taking a trigonometry class. I believe i learned this in algebra but i can't remember how its done.

    cos = +/-√1-225/289

    (basically i don't understand how to do the square root part of this. its the square root of 1- 225 over 289.

    The answer for this is = +/- 8/17

    **no clue how they got 8/17.
    thx and godbless if anybody can help me

    http://i48.photobucket.com/albums/f2...799/sqroot.jpg
    <br />
\sqrt{1 - \frac{225}{289}}<br />
\Rightarrow \sqrt{\frac{289}{289} - \frac{225}{289}}<br />
\Rightarrow \sqrt{\frac{64}{289}}<br />
\Rightarrow \pm\frac{8}{17}<br />

    \cos{\theta} = \pm\sqrt{1-\frac{225}{289}}
    square both sides
    \cos^2{\theta} =  1 - \frac{225}{289}<br />
\Rightarrow<br />
\cos^2{\theta}-1 =  \frac{225}{289}<br />
\Rightarrow<br />
\sin^2{\theta} = \sqrt{\frac{225}{298}}<br />
\Rightarrow<br />
\sin{\theta = \pm\frac{15}{17}

    do you need help with more of this
    Last edited by bigwave; June 23rd 2010 at 08:20 PM. Reason: more steps
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