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Math Help - trigonometric identity how should i start this

  1. #1
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    trigonometric identity how should i start this

    sin theta/ sin theta - cos theta

    = (convert to)

    sin theta (sin theta - cos theta)/ 1 - 2 sin theta cos theta

    how should i start this?
    pls.. help
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  2. #2
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    Quote Originally Posted by xahjika View Post
    sin theta/ sin theta - cos theta

    = (convert to)

    sin theta (sin theta - cos theta)/ 1 - 2 sin theta cos theta

    how should i start this?
    pls.. help
    Multiply \frac{\sin{\theta}}{\sin{\theta}-\cos{\theta}} by \frac{\sin{\theta}-\cos{\theta}}{\sin{\theta}-\cos{\theta}}

    This one should have been very obvious.
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  3. #3
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    ok tnks.. haha..

    by the way is this right?

    = sin theta (sin theta - cos theta)/ (sin theta - cos theta)(sin theta - cos theta)

    = sin theta (sin theta - cos theta)/ sin^2 theta - sin theta cos theta - cos theta sin theta + cos^2 theta

    (sin^2 theta + cos^2 theta = 1)
    (- sin theta cos theta - cos theta sin theta= -2 sin theta cos theta)

    = sin theta (sin theta - cos theta)/ 1-2 sin theta cos theta
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  4. #4
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    Quote Originally Posted by xahjika View Post
    ok tnks.. haha..

    by the way is this right?

    = sin theta (sin theta - cos theta)/ (sin theta - cos theta)(sin theta - cos theta)

    = sin theta (sin theta - cos theta)/ sin^2 theta - sin theta cos theta - cos theta sin theta + cos^2 theta

    (sin^2 theta + cos^2 theta = 1)
    (- sin theta cos theta - cos theta sin theta= -2 sin theta cos theta)

    = sin theta (sin theta - cos theta)/ 1-2 sin theta cos theta
    Bravo.
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  5. #5
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    Re:

    \frac{sin\Theta}{sin\Theta-cos\Theta}=\frac{sin\Theta(sin\Theta-cos\Theta)}{1-2sin\Theta cos\Theta}

    I assume that you mean this .

    From RHS ,
    \frac{sin\Theta(sin\Theta-cos\Theta)}{cos^2\Theta-2sin\Theta cos\Theta+sin^2\Theta}
    = \frac{sin\Theta(sin\Theta-cos\Theta)}{(cos\Theta-sin\Theta)^2}
    = \frac{-sin\Theta(cos\Theta-sin\Theta)}{(cos\Theta-sin\Theta)^2}
    = \frac{-sin\Theta}{-(sin\Theta-cos\Theta)}
    = \frac{sin\Theta}{sin\Theta-cos\Theta}
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