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Math Help - ~Trigonometry

  1. #1
    Senior Member slevvio's Avatar
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    ~Trigonometry

    Prove this identity:

     \frac{cos2A}{1+sin2A} = tan(\frac{\pi}{4} - A)

    Thanks I was having trouble proving this!
    In the previous question I was able to prove:

     \frac{sin2A}{1 + cos2A} = tanA
    Last edited by slevvio; November 7th 2007 at 01:33 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by slevvio View Post
    Prove this identity:

     \frac{cos2A}{1+sin2A} = tan(\frac{\pi}{2} - A)

    Thanks I was having trouble proving this!
    I'm not surprised you are having trouble proving it. It isn't true.

    -Dan
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  3. #3
    Senior Member slevvio's Avatar
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    Oh dear, I copied the angle down wrong ! I have edited my post
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by slevvio View Post
    Prove this identity:

     \frac{cos2A}{1+sin2A} = tan(\frac{\pi}{4} - A)
    That's better.

    Let's do this one from the RHS since I don't know which form of cos(2A) will be the most helpful.

    tan \left ( \frac{\pi}{4} - A \right ) = \frac{tan \left ( \frac{\pi}{4} \right ) - tan(A)}{1 + tan \left ( \frac{\pi}{4} \right ) \cdot tan(A)}

    = \frac{1 - tan(A)}{1 + tan(A)}

    = \frac{1 - \frac{sin(A)}{cos(A)}}{1 + \frac{sin(A)}{cos(A)}} <-- Clear the complex fraction

    = \frac{cos(A) - sin(A)}{cos(A) + sin(A)}

    Now I'm going to simplify by multiplying top and bottom by cos(A) + sin(A):
    = \frac{cos^2(A) - sin^2(A)}{cos^2(A) +  2sin(A)cos(A) + sin^2(A)}

    = \frac{cos(2A)}{1 + sin(2A)}

    Can I cook or can I cook?

    -Dan
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  5. #5
    Senior Member slevvio's Avatar
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    You certainly can sautée!
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