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  1. #1
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    identity

    Prove that cos2x= 1-tan^2x/ 1+tan^2x

    Thanks =)
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Hellooo View Post
    Prove that cos2x= 1-tan^2x/ 1+tan^2x

    Thanks =)
    RHS = \frac{1-tan^{2}x}{1+tan^{2}x} = \frac{1 - \frac{sin^{2}x}{cos^{2}x}}{1+\frac{sin^{2}x}{cos^{  2}x}}

    try finishing it now
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  3. #3
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    Hello, Hellooo!

    Going from left to right is tricky . . .


    Prove that: . \cos2x\:=\:\frac{1-\tan^2\!x}{1+\tan^2\!x}

    We have: . \cos2x \;=\;\frac{\cos^2\!x - \sin^2\!x}{1}

    Divide top and bottom by \cos^2\!x\!:\quad \frac{\dfrac{\cos^2\!x}{\cos^2\!x} - \dfrac{\sin^2\!x}{\cos^2\!x}}{\dfrac{1}{\cos^2\!x}  }

    . . and we have: . \frac{1-\tan^2\!x}{\sec^2\!x} \;=\;\frac{1-\tan^2\!x}{1+\tan^2\!x}

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  4. #4
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    Thnx for the quick replies!!

    if i didn't know that sec^2x =1+ tan^2x, is it a lot harder to prove that 1-tan^2x/1+tan^2x = cos2x?

    i just did this in a test, and couldn't figure it out because i didn't memorize the sec, cot, or csc identities >.<
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  5. #5
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    Quote Originally Posted by Hellooo View Post
    Thnx for the quick replies!!

    if i didn't know that sec^2x =1+ tan^2x, is it a lot harder to prove that 1-tan^2x/1+tan^2x = cos2x?
    Not at all.
    Take
    (1-tan^2x)/(1+tan^2x)
    Multiply and divide by cos^2x. You get
    [cos^2x - tan^2x*cos^2x]/[cos^2x + tan^2x*cos^2x]
    I hope, you know that tanx = sinx/cosx and basic indentity cos^2x + sin^2x = 1
    So you get
    [cos^2x - sin^2x]/[cos^2x + sin^2x]
    = [cos^2x - sin^2x] = cos(2x)
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  6. #6
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    oh thnx a lot!! =))
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