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Math Help - Mathematics - Really Simple Trig Identities that I can't get!

  1. #1
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    Mathematics - Really Simple Trig Identities that I can't get!

    Prove:
    (tanx)(sinx) / (tanx) + (sinx) = (tanx) - (sinx) / (tanx)(sinx)


    What I have so far:

    L.S.
    = (sinx / cosx) sinx / (sinx / cosx) + sinx
    = (sin^2x / cosx) / (sinx + (sinx) (cosx) / cosx)
    = (sin^2x / cosx) / (cosx / sinx + sinxcosx)

    What do I do now and if I did something wrong, where did I go wrong and how should I go about to fix it?
    Last edited by Macleef; November 11th 2007 at 03:41 AM.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Macleef View Post
    Prove:
    sin^2x - sin^4x = cos^2x - cos^4x

    What I have:

    L.S.
    = sin^2x - sin^4x
    = sin^2x (-sin^2x)
    = 1 - cos^2x (-1 + cos^2x)
    = -1 +cos^2x + cos^2x - cos^4x
    = cos^2x - 1 +cos^2x - cos^4x
    = sin^2x + cos^2x - cos^4x

    Where did I go wrong?
    sin^2x - sin^4x = sin^2x (1 - sin^2 x) = (1 - cos^2 x) cos^2 x = RHS
    ---

    Quote Originally Posted by Macleef View Post
    Prove: sinx - 1 / sinx + 1 = - cos^2x / (sinx + 1)^2

    What I have so far:

    R.S.
    = - (1 - sin^2x) / (sinx + 1) (sinx + 1)
    = -1 + sin^2x / (sinx + 1) (sinx + 1)

    What do I do now?
    \frac{- cos^2x }{(sinx + 1)^2} = \frac{-(1 - sin^2 x)}{(sinx + 1)^2}= \frac{-(1 - sinx)(1 + sinx)}{(sinx + 1)^2}
    can you continue now?
    ---
    Quote Originally Posted by Macleef View Post
    Prove:
    (tanx)(sinx) / (tanx) + (sinx) = (tanx) - (sinx) / (tanx)(sinx)

    What I have so far:

    L.S.
    = (sinx / cosx) sinx / (sinx / cosx) + sinx
    = (sin^2x / cosx) / (sinx + (sinx) (cosx) / cosx)
    = (sin^2x / cosx) / (cosx / sinx + sinxcosx)

    What do I do now and if I did something wrong, where did I go wrong and how should I go about to fix it?
    LHS = \frac{\frac{sin^2x}{cosx}}{\frac{sinx}{cosx}+sinx} = \frac{\frac{sin^2x}{cosx}}{\frac{sinx + sinx cosx}{cosx}}

    = \frac{sin^2x}{sinx + sinx cosx} = \frac{1 - cos^2x}{sinx(1 + cosx)} = \frac{(1 - cosx)(1 + cosx)}{sinx(1 + cosx)}

    \frac{1 - cosx}{sinx}, multiplying \frac{tanx}{tanx} yields RHS..
    Last edited by kalagota; November 10th 2007 at 04:17 AM.
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  3. #3
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    Prove:
    sin^2x - sin^4x = cos^2x - cos^4x

    sin^2x - sin^4x = sin^2x (1 - sin^2 x) = (1 - cos^2 x) cos^2 x = RHS


    ^^^

    How did you get "sin^2x(1 - sin^2 x)" in the second line? When I did this, I got "sin^2x (-sin^2x)"
    Can you tell me what I did wrong ? I don't know how you got the "1" from ...


    ---


    \frac{- cos^2x }{(sinx + 1)^2} =  \frac{-(1 - sin^2 x)}{(sinx + 1)^2}= \frac{-(1 - sinx)(1 + sinx)}{(sinx + 1)^2}
    can you continue now?

    When I did this, I get:
    = (-1 + sinx) (1 + sinx) / (sinx + 1) (sinx +1)
    = (-1 + sinx) / (sinx + 1)

    And now I'm stumped? I don't know what to do ...
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Macleef View Post
    How did you get "sin^2x(1 - sin^2 x)" in the second line? When I did this, I got "sin^2x (-sin^2x)"
    Can you tell me what I did wrong ? I don't know how you got the "1" from ...
    ---
    remember factoring.. (a+ab)=a(1+b), so when you factored out sin^2 x from sin^2 x - sin^4 x, it must be sin^2 x (1 - sin^2 x).. you should remain 1 from the parenthesis..

    Quote Originally Posted by Macleef View Post
    When I did this, I get:
    = (-1 + sinx) (1 + sinx) / (sinx + 1) (sinx +1)
    = (-1 + sinx) / (sinx + 1)

    And now I'm stumped? I don't know what to do ...
    you are done.. that's what you want to show right?
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  5. #5
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    ..

    Quote Originally Posted by kalagota View Post
    ---
    remember factoring.. (a+ab)=a(1+b), so when you factored out sin^2 x from sin^2 x - sin^4 x, it must be sin^2 x (1 - sin^2 x).. you should remain 1 from the parenthesis..



    you are done.. that's what you want to show right?

    The first one, I remember it now, thanks

    And as for the second one,

    I'm proving

    sinx - 1 / six + 1 = -cos^2x / (sinx + 1)^2

    But ... when I did RS = RS, I get this:

    = (-1 + sinx) / (sinx + 1)
    = - (1 - sinx) / (sinx + 1)

    Why do I have a negative in the numerator next to the bracket? How do I remove this to prove the identity?
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  6. #6
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Macleef View Post
    The first one, I remember it now, thanks

    And as for the second one,

    I'm proving

    sinx - 1 / six + 1 = -cos^2x / (sinx + 1)^2

    But ... when I did RS = RS, I get this:

    = (-1 + sinx) / (sinx + 1)
    = - (1 - sinx) / (sinx + 1)

    Why do I have a negative in the numerator next to the bracket? How do I remove this to prove the identity?
    wait, did u mean like this?

    sinx - \frac{1}{sinx} + 1 or  sinx - \frac{1}{sinx + 1} or  \frac{sinx - 1}{sinx + 1}

    first, second or third?
    Last edited by kalagota; November 10th 2007 at 05:11 AM.
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  7. #7
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    Quote Originally Posted by kalagota View Post
    wait, did u mean like this?

    sinx - \frac{1}{sinx} + 1 or  sinx - \frac{1}{sinx + 1} or  \frac{sinx - 1}{sinx + 1}

    first, secodn or third?
    I mean like

    = - (1 - sinx) over (sinx + 1)

    or

    - (1 - sinx)
    _____________
    sinx + 1
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  8. #8
    MHF Contributor kalagota's Avatar
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    i am asking what is your left hand side.. first second or third?

    coz if its the third one, then you are done with

    \frac{-1 + sinx}{sinx + 1}..
    remember that addition of real number is commutative so
    \frac{-1 + sinx}{sinx + 1} = \frac{sinx - 1}{sinx + 1} which is the third one..
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  9. #9
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    wow I see it now, thanks, I feel so stupid right now -_-
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