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Math Help - Trig Identity Proof

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    Trig Identity Proof

    Prove the identity 1 - cos\;2\;\theta = \frac{2}{1 + cot^2\;\theta}

    I'm a little lost where to begin on this one. Does anyone have any tips for proving these identities?
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  2. #2
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    Quote Originally Posted by TI-84 View Post
    Prove the identity 1 - cos\;2\;\theta = \frac{2}{1 + cot^2\;\theta}

    I'm a little lost where to begin on this one. Does anyone have any tips for proving these identities?
    R.H.S. = \frac{2}{1 + \cot^2 \theta} = \frac{2}{1 + \frac{\cos^2 \theta}{\sin^2 \theta}} = \frac{2 \sin^2 \theta}{\sin^2 \theta + \cos^2 \theta} = 2 \sin^2 \theta

    = -(1 - 2 \sin^2 \theta) + 1 = -\cos(2 \theta) + 1 =L.H.S.

    I knew to do the last couple of bits because L.H.S. =1 - \cos (2 \theta) = 1 - (\cos^2 \theta - \sin^2 \theta) = 1 - \cos^2 \theta + \sin^2 \theta = 2 \sin^2 \theta.
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    I did 1 - cos\;2\theta = \frac{2}{csc^2\;\theta}

    1 - cos\;2\theta = 2sin^2\;\theta

    1 - cos\;2\theta = 2(1 - cos^2\;\theta)

    1 - cos\;2\theta = 2 - 2cos^2\;\theta

    1 - (1 - 2sin^2\;\theta) = 2 - 2(1 - sin^2\;\theta)

    2sin^2\;\theta = 2 - 2 + 2sin^2\;\theta

    2sin^2\;\theta = 2sin^2\;\theta

    Hows it look compared to yours?
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    Quote Originally Posted by TI-84 View Post
    I did 1 - cos\;2\theta = \frac{2}{csc^2\;\theta}

    1 - cos\;2\theta = 2sin^2\;\theta

    1 - cos\;2\theta = 2(1 - cos^2\;\theta)

    1 - cos\;2\theta = 2 - 2cos^2\;\theta

    1 - (1 - 2sin^2\;\theta) = 2 - 2(1 - sin^2\;\theta)

    2sin^2\;\theta = 2 - 2 + 2sin^2\;\theta

    2sin^2\;\theta = 2sin^2\;\theta

    Hows it look compared to yours?
    Well, that's a much better way.

    I'd make just a couple of small changes so that the solution agrees with the standard setting out of an identity proof:

    R.H.S. = \frac{2}{csc^2\;\theta}

    = 2 \sin^2\;\theta

     = 2(1 - \cos^2\;\theta)

     = 2 - 2 \cos^2\;\theta And here's where we reach the fork in the road because you need to keep working with the R.H.S.

     = 1 + (1 - 2 \cos^2\;\theta)

    = 1 - (2 \cos^2\;\theta - 1)

    = 1 - \cos (2 \theta )

    = L.H.S.

    Do you see the style required?
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    Question. Can you work with both sides when proving an identity, as long as they don't interfere with one another.
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  6. #6
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    Quote Originally Posted by TI-84 View Post
    Question. Can you work with both sides when proving an identity, as long as they don't interfere with one another.
    Well, I guess you can ..... but it's not good form in my opinion. Usually if you want to do that, you can instead use it to see how the LHS (or RHS) needs to continue.

    In my opinion, trig identities should follow the format:

    This side (LHS or RHS as required) = whatever it is

    = blah blah

    = yada yada

    = ho hum ho hum
    .
    .
    .

    = the other side (RHS or LHS as required).
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