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Thread: proving trig identity

  1. #1
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    proving trig identity

    This has been alluding me for a week and I am now down to the wire

    Prove that (sin^2(x)) / (1-cos(x)) equals (sec(x)+1)/(sec(x))


    any help,hints,clues or outright answers would be appreciated!

    Thanks,
    mj
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by maryjane View Post
    This has been alluding me for a week and I am now down to the wire

    Prove that (sin^2(x)) / (1-cos(x)) equals (sec(x)+1)/(sec(x))


    any help,hints,clues or outright answers would be appreciated!

    Thanks,
    mj
    Consider the $\displaystyle LHS$

    $\displaystyle \frac {\sin^2 x}{1 - \cos x} = \frac {1 - \cos^2 x}{1 - \cos x}$ ...........Pythagorean identity

    $\displaystyle = \frac {(1 + \cos x )( 1 - \cos x)}{1 - \cos x}$ ..........Numerator was the difference of two squares

    $\displaystyle = 1 + \cos x$ ..........................cancel

    $\displaystyle = \frac {\sec x}{\sec x} + \frac 1{\sec x}$ .....................algebraic manipulation

    $\displaystyle = \frac {\sec x + 1}{\sec x}$ ...........................combine the fractions

    $\displaystyle = RHS$

    Thus, $\displaystyle LHS \equiv RHS$, as desired
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by maryjane View Post
    This has been alluding me for a week and I am now down to the wire

    Prove that (sin^2(x)) / (1-cos(x)) equals (sec(x)+1)/(sec(x))


    any help,hints,clues or outright answers would be appreciated!

    Thanks,
    mj
    $\displaystyle \frac{\sin^2(x)}{1-\cos(x)}=\frac{\sec(x)+1}{\sec(x)}$

    we know that $\displaystyle \sin^2(x)=1-\cos^2(x)$

    and we also know that $\displaystyle a^2-b^2=(a+b)(a-b)$

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

    So

    now subbing we get

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

    Cancelling the left side and splitting the left we get

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

    now we know that $\displaystyle \frac{\sec(x)}{\sec(x)}=1$

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

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

    imputting back in gives

    $\displaystyle 1+\cos(x)=1+\cos(x)$
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  4. #4
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    well, well, well.... I now feel like an idiot
    thank-you both!

    mj
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by maryjane View Post
    well, well, well.... I now feel like an idiot
    thank-you both!

    mj
    *spider-man voice*Mary Jane! hang On!

    you're welcome ..........that's Kriz's icon
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  6. #6
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    haha

    one more question I have...how do we get

    sec(x) + 1
    sec(x) sec(x)

    from the 1 + cos(x) ... I realize we can sub the first part for the 1
    but how do we sub 1
    sec(x) from cos(x) ?

    I am obviously missing something (i wish i could find it)

    mj
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by maryjane View Post
    haha

    one more question I have...how do we get

    sec(x) + 1
    sec(x) sec(x)

    from the 1 + cos(x) ... I realize we can sub the first part for the 1
    but how do we sub 1
    sec(x) from cos(x) ?

    I am obviously missing something (i wish i could find it)

    mj
    If you are takling about how $\displaystyle \frac{\sec(x)+1}{\sec(x)}=1+\cos(x)$

    its because $\displaystyle \frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$

    the rest I showed in my repsonse
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