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Math Help - Proving Trigonometric Identities

  1. #1
    Super Member Quacky's Avatar
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    Proving Trigonometric Identities

    This question seems really easy, but I can't seem to work it out.

    Prove that:
    EDITTED: SEE BELOW

    These are the identities we can use:
    Cos^2\theta+Sin^2\theta=1
    \frac{Sin\theta}{cos\theta}=tan\theta

     1+tan^2\theta=sec^2\theta
     cosec^2\theta=1+cot^2\theta

    Thanks in advance for any help, please check back after posting as I may have a few problems.

    Apologies, I've worked it out by myself. I'll edit the post if I have any other issues. Sorry for wasting anyone's time.
    Last edited by Quacky; January 20th 2010 at 03:21 PM.
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  2. #2
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    1+tan^2\theta=1+\frac{sin^2\theta}{cos^2\theta}=\f  rac{cos^2\theta}{cos^2\theta}+\frac{sin^2\theta}{c  os^2\theta}=\frac{sin^2\theta+cos^2\theta}{cos^2\t  heta}

    1+cot^2\theta=\frac{sin^2\theta}{sin^2\theta}+\fra  c{cos^2\theta}{sin^2\theta}
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  3. #3
    Super Member Quacky's Avatar
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    Sorry Archie , Thanks for confirming my answer at least.
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  4. #4
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    Hello, Quacky!

    Prove that: . \cot^2\theta + \cos^2\theta \;=\;(\csc \theta - \sin\theta)(\csc\theta + \sin\theta)
    I started with the right side . . .

    (\csc\theta -\sin\theta)(\csc\theta + \sin\theta) \;\;=\;\;\csc^2\!\theta - \sin^2\theta \;\;=\;\;\frac{1}{\sin^2\!\theta} - \sin^2\theta

    . . =\;\; \frac{1-\sin^4\!\theta}{\sin^2\!\theta} \;\;=\;\;\frac{(1-\sin^2\!\theta)(1 + \sin^2\!\theta)}{\sin^2\!\theta} \;\;=\;\;\frac{\cos^2\!\theta(1 + \sin^2\!\theta)}{\sin^2\!\theta}

    . . =\;\; \frac{\cos^2\!\theta + \sin^2\!\theta\cos^2\!\theta}{\sin^2\!\theta} \;\;=\;\;\frac{\cos^2\!\theta}{\sin^2\!\theta} + \frac{\sin^2\!\theta\cos^2\!\theta}{\sin^2\!\theta  }

    . . =\;\; \left(\frac{\cos\theta}{\sin\theta}\right)^2 + \cos^2\!\theta \;\;=\;\;\cot^2\!\theta + \cos^2\!\theta

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  5. #5
    Super Member Quacky's Avatar
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    I feel really bad now , thanks though.
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  6. #6
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    Don't feel bad, Quacky,
    go easy on yourself when you are learning,
    it's a delicate time to go through.
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