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Thread: proving trigonometric identity

  1. #1
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    proving trigonometric identity

    Write the following expression in either sine or cosine

    $\displaystyle 1) sin^2x+ cos^2x * cscx $

    Attempt:
    $\displaystyle sin^2 + cos^2 * \frac{1}{sinx} $
    $\displaystyle sin^2 + \frac{cos^2}{sinx}$
    $\displaystyle \frac{sin^2x}{1} + \frac{cos^2}{sinx}$
    $\displaystyle \frac{sin^3x+cos^2}{sinx} $
    $\displaystyle sin^3 +cos^2 * \frac{1}{sinx}$
    $\displaystyle 1*1=1 $


    Can you please answer this question too
    $\displaystyle cosx +sin^2x* secx $

    Are my above steps correct please write in step if they are incorrect.
    Thank you very much!
    Last edited by mj.alawami; Feb 16th 2009 at 02:55 AM.
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  2. #2
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    Quote Originally Posted by mj.alawami View Post
    Write the following expression in either sine or cosine

    $\displaystyle 1) sin^2x+ cos^2x * cscx $

    Attempt:
    $\displaystyle sin^2 + cos^2 * \frac{1}{sinx} $
    $\displaystyle sin^2 + \frac{cos^2}{sinx}$
    $\displaystyle \frac{sin^2x}{1} + \frac{cos^2}{sinx}$
    $\displaystyle \frac{sin^3x+cos^2}{sinx} $
    $\displaystyle sin^3 +cos^2 * \frac{1}{sinx}$
    $\displaystyle 1*1=1 $


    $\displaystyle \frac{sin^3x+cos^2}{sinx} $

    $\displaystyle
    =\frac{sin^3x-sin^2x+1}{sinx}
    $
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  3. #3
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by mj.alawami View Post
    Write the following expression in either sine or cosine

    $\displaystyle 1) sin^2x+ cos^2x * cscx $

    Attempt:
    $\displaystyle sin^2 + cos^2 * \frac{1}{sinx} $
    $\displaystyle sin^2 + \frac{cos^2}{sinx}$
    $\displaystyle \frac{sin^2x}{1} + \frac{cos^2}{sinx}$
    $\displaystyle \frac{sin^3x+cos^2}{sinx} $
    _______________________________________
    $\displaystyle (sin^3 +cos^2 )* \frac{1}{sinx}$
    _______________________________________
    ___________________________
    $\displaystyle 1*1=1 $
    __________________________

    Can you please answer this question too
    $\displaystyle cosx +sin^2x* secx $

    Are my above steps correct please write in step if they are incorrect.
    Thank you very much!
    -First of all watch the step in Red (correct)

    -Now watch the step in Green (incorrect)

    If your question is (doubtful since not a very good use of bracket)

    $\displaystyle cosx +sin^2x* secx $


    $\displaystyle = \frac{sin^2(x) }{cos(x)} +cos(x) $


    $\displaystyle
    = \frac{sin^2(x) +cos^2(x)}{cos(x)}

    $

    Since
    $\displaystyle
    sin^2(x) +cos^2(x) =1$

    $\displaystyle
    = \frac{1}{cos(x)}
    $
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