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Math Help - Trig Identities

  1. #1
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    Trig Identities

    Prove that
    1-sin^2X/1+cotX - cos^2X/1+tanX = sinXcosX

    Much thanks in advance.
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  2. #2
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    Quote Originally Posted by ahling View Post
    Prove that
    1-sin^2X/1+cotX - cos^2X/1+tanX = sinXcosX

    Much thanks in advance.
    Please prove that you can add parentheses to clarify meaning.

    Notice:

    1 + x / x - 1 = 1 + (x / x) - 1 = 1 + (1) - 1 = 2 - 1 = 1

    However,

    (1 + x) / (x - 1) is a very different animal.
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  3. #3
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    Yes, I understand, but .. in the original equation, there are no parantheses. I typed the question exactly as it is shown on the paper given to me.
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  4. #4
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    Hello, ahling!

    Prove that: . 1 - \frac{\sin^2\!x}{1 + \cot x}- \frac{\cos^2\!x}{1 + \tan x} \:=\:\sin x\cos x
    We have: . 1 - \frac{\sin^2\!x}{1 + \dfrac{\cos x}{\sin x}} - \frac{\cos^2\!x}{1 + \dfrac{\sin x}{\cos x}}


    Multiply the first fraction by \frac{\sin x}{\sin x} , the second by \frac{\cos x}{\cos x}

    . . 1 - \frac{\sin^3\!x}{\sin x + \cos x} - \frac{\cos^3\!x}{\cos x + \sin x} \;\;=\;\;1 - \frac{sin^3\!x + \cos^3\!x}{\sin x + \cos x}


    . . = \;\;1 - \frac{(\sin x + \cos x)(\sin^2\!x - \sin x\cos x + \cos^2\!x)}{\sin x + \cos x}


    . . = \;\;1 - (\underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}} - \sin x\cos x) \;\;=\;\;1 - (1 - \sin x\cos x)


    . . = \;\sin x\cos x . . . ta-DAA!



    And LateX is working !!

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  5. #5
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    Quote Originally Posted by ahling View Post
    in the original equation, there are no parantheses. I typed the question exactly as it is shown on the paper given to me.
    I have a little trouble believing that, but let's go with it...

    1-sin^2X/1+cotX - cos^2X/1+tanX = sinXcosX

    The "/1" are meaningless in this form, so...

    1-sin^2X+cotX - cos^2X+tanX = sinXcosX

    1-sin^2X = cos^2X <== Identity to Substitute

    cos^2X+cotX - cos^2X+tanX = sinXcosX

    cotX+tanX = sinXcosX

    Since cotX = (cosX)/(sinX) and tanX = (sinX)/(cosX)
    And Since cos^2X + sin^2X = 1

    1/(sinXcosX) = sinXcosX

    I think you would have a hard time selling that as an identity.

    Note: I would do this on an exam. It is technically correct, but may have points taken off for being a smart aleck. I would also rewrite it as it probably is intended and work it that way. This may avoid the smart aleck points. Don't put up with sloppy notation. Certainly don't write any yourself. Everyone needs reminders to be clear - particularly if they want to be understood generally.

    LaTeX Check: \cos^{2}(x) + \sin^{2}(x) = 1
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