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Math Help - Proving Sum and difference identities

  1. #1
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    Proving Sum and difference identities

    Question:
     \frac{sin(u+v)+sin(u-v)}{cos(u+v)+cos(u-v)} =tan u

    Attempt:
    I simplifyied using the equation and i reached to a point where my answer is
     tan(u+v) + tan(u-v)

    I have no idea what to do next to prove...




    Thank you..
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  2. #2
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    Quote Originally Posted by mj.alawami View Post
    Question:
     \frac{sin(u+v)+sin(u-v)}{cos(u+v)+cos(u-v)} =tan u

    Attempt:
    I simplifyied using the equation and i reached to a point where my answer is
     tan(u+v) + tan(u-v)

    I have no idea what to do next to prove...

    Thank you..
    \sin({u+v}) = sin(u)cos(v) + sin(v)cos(u)

    \sin({u-v}) = sin(u)cos(v) - sin(v)cos(u)

    \cos({u+v}) = cos(u)cos(v) - sin(v)sin(u)

    \cos({u-v}) = cos(u)cos(v) + sin(v)sin(u)

    so we get:

    \frac{sin(u)cos(v) + sin(v)cos(u) + sin(u)cos(v) - sin(v)cos(u)}{cos(u)cos(v) - sin(u)sin(v) + cos(u)cos(v) + sin(u)sin(v)}

    Factorising gives us

    \frac{sin(u)(2cos(v)) + sin(v)(0)}{cos(u)(2cos(v)) + sin(v)(0)}

    Cancelling 2cos(v) and because anything times 0 is 0 we get:

    \frac{sin(u)}{cos(u)} = tan(u)
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    \sin({u+v}) = sin(u)cos(v) + sin(v)cos(u)

    \sin({u-v}) = sin(u)cos(v) - sin(v)cos(u)

    \cos({u+v}) = cos(u)cos(v) - sin(v)sin(u)

    \cos({u-v}) = cos(u)cos(v) + sin(v)sin(u)

    so we get:

    \frac{sin(u)cos(v) + sin(v)cos(u) + sin(u)cos(v) - sin(v)cos(u)}{cos(u)cos(v) - sin(u)sin(v) + cos(u)cos(v) + sin(u)sin(v)}

    Factorising gives us

    \frac{sin(u)(2cos(v)) + sin(v)(0)}{cos(u)(2cos(v)) + sin(v)(0)}

    Cancelling 2cos(v) and because anything times 0 is 0 we get:

    \frac{sin(u)}{cos(u)} = tan(u)

    How did you factorize it because i am getting lost?
    Last edited by mj.alawami; February 28th 2009 at 08:32 AM.
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    Quote Originally Posted by mj.alawami View Post
    Question:
     \frac{sin(u+v)+sin(u-v)}{cos(u+v)+cos(u-v)} =tan u
    \frac{\sin(u+v)+\sin(u-v)}{\cos(u+v)+\cos(u-v)}

    =\frac{(\sin u\cos v+\cos u\sin v)+(\sin u\cos v-\cos u\sin v)}{(\cos u\cos v-\sin u\sin v)+(\cos u\cos v+\sin u\sin v)}

    =\frac{2\sin u\cos v}{2\cos u\cos v}

    =\frac{\sin u}{\cos u}=\tan u.

    Attempt:
    I simplifyied using the equation and i reached to a point where my answer is
     tan(u+v) + tan(u-v)

    I have no idea what to do next to prove...
    This is not correct. For example,

    \tan\left(\frac\pi6+\frac\pi6\right)+\tan\left(\fr  ac\pi6-\frac\pi6\right)=\sqrt3\neq\frac{\sqrt3}3=\tan\fra  c\pi6.

    Can you show your work?
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by mj.alawami View Post
    How did you factorize it ?
    It came from combining like terms, I could have rearranged them first in an extra step which would be going from:

    \frac{sin(u)cos(v) + sin(v)cos(u) + sin(u)cos(v) - sin(v)cos(u)}{cos(u)cos(v) - sin(u)sin(v) + cos(u)cos(v) + sin(u)sin(v)}

    to

    \frac{(sin(u)cos(v) + sin(u)cos(v))+ (sin(v)cos(u) - sin(v)cos(u))}{(cos(u)cos(v) + cos(u)cos(v)) - (sin(u)sin(v) + sin(u)sin(v))}

    I could also have then simplified each set of brackets to give:


    (sin(u)cos(v) + sin(u)cos(v)) = 2sin(u)cos(v)

    (sin(v)cos(u) - sin(v)cos(u)) = 0

    (cos(u)cos(v) + cos(u)cos(v)) = 2cos(u)cos(v)

    (-sin(u)sin(v) + sin(u)sin(v)) = 0

    this means the 'main' equation will be :

    \frac{2sin(u)cos(v)}{2cos(u)cos(v)}

    As 2cos(v) is on the top and bottom it can be cancelled to give {sin(u)}{cos(u)} = tan(u)
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  6. #6
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    Quote Originally Posted by e^(i*pi) View Post
    It came from combining like terms, I could have rearranged them first in an extra step which would be going from:

    \frac{sin(u)cos(v) + sin(v)cos(u) + sin(u)cos(v) - sin(v)cos(u)}{cos(u)cos(v) - sin(u)sin(v) + cos(u)cos(v) + sin(u)sin(v)}

    to

    \frac{(sin(u)cos(v) + sin(u)cos(v))+ (sin(v)cos(u) - sin(v)cos(u))}{(cos(u)cos(v) + cos(u)cos(v)) - (sin(u)sin(v) + sin(u)sin(v))}

    I could also have then simplified each set of brackets to give:


    (sin(u)cos(v) + sin(u)cos(v)) = 2sin(u)cos(v)

    (sin(v)cos(u) - sin(v)cos(u)) = 0

    (cos(u)cos(v) + cos(u)cos(v)) = 2cos(u)cos(v)

    (-sin(u)sin(v) + sin(u)sin(v)) = 0

    this means the 'main' equation will be :

    \frac{2sin(u)cos(v)}{2cos(u)cos(v)}

    As 2cos(v) is on the top and bottom it can be cancelled to give {sin(u)}{cos(u)} = tan(u)
    Thanks alott it sunk in very easily
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