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Math Help - [SOLVED] Opposite Angle Identities

  1. #1
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    [SOLVED] Opposite Angle Identities

    Hi, I am working on the following problem and do not understand why the final answer is slightly different than my answer:

    If cos t = ( - 1/3 ) and ( (pi/2) < t < pi )
    Compute sin ( -t ) + cos ( -t )


    I find sin t to be ( 2 (sqrt 2) / 3 ) using the pythag. theorem.

    sin ( -t ) = - sin t... this makes 2 (sqrt 2) / 3 a negative value
    cos ( -t ) = cos t... this leaves the negative value of ( - 1/3 )

    I then combine over the denominator of 3:

    ( - 2 (sqrt 2) - 1 ) / 3

    The answer in the book is - (( 2 (sqrt 2) + 1) / 3 )

    I know if I distribute the negative, I'll end up with the same answer, but why is the negative pulled to the front of the answer?

    Thanks!
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by Snowcrash View Post
    Hi, I am working on the following problem and do not understand why the final answer is slightly different than my answer:

    If cos t = ( - 1/3 ) and ( (pi/2) < t < pi )
    Compute sin ( -t ) + cos ( -t )

    I find sin t to be ( 2 (sqrt 2) / 3 ) using the pythag. theorem.

    sin ( -t ) = - sin t... this makes 2 (sqrt 2) / 3 a negative value
    cos ( -t ) = cos t... this leaves the negative value of ( - 1/3 )

    I then combine over the denominator of 3:

    ( - 2 (sqrt 2) - 1 ) / 3

    The answer in the book is - (( 2 (sqrt 2) + 1) / 3 )

    I know if I distribute the negative, I'll end up with the same answer, but why is the negative pulled to the front of the answer?

    Thanks!
    Hi Snowcrash,

    That's just another way of expressing the same result.

    \frac{-2\sqrt{2}-1}{3}=-\left(\frac{2\sqrt{2}+1}{3}\right)
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  3. #3
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    Quote Originally Posted by Snowcrash View Post
    Hi, I am working on the following problem and do not understand why the final answer is slightly different than my answer:

    If cos t = ( - 1/3 ) and ( (pi/2) < t < pi )
    Compute sin ( -t ) + cos ( -t )


    I find sin t to be ( 2 (sqrt 2) / 3 ) using the pythag. theorem.

    sin ( -t ) = - sin t... this makes 2 (sqrt 2) / 3 a negative value
    cos ( -t ) = cos t... this leaves the negative value of ( - 1/3 )

    I then combine over the denominator of 3:

    ( - 2 (sqrt 2) - 1 ) / 3

    The answer in the book is - (( 2 (sqrt 2) + 1) / 3 )

    I know if I distribute the negative, I'll end up with the same answer, but why is the negative pulled to the front of the answer?

    no reason other than to make it look nice, that's all.

    \textcolor{red}{\frac{-a-b}{c} = -\frac{a+b}{c}}
    ...
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  4. #4
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    Great, that's exactly what I wanted to hear - thank you!
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