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Math Help - Another Identity

  1. #1
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    Another Identity

    I have gotten stuck again on another problem.

    [LaTeX ERROR: Convert failed]

    So this is what I did was made one fraction in the denominator:

    [LaTeX ERROR: Convert failed]

    So then I multiplied by the recip. and simplified and worked with the other fraction to get:\
    [LaTeX ERROR: Convert failed]

    So I can do this
    [LaTeX ERROR: Convert failed]

    I do not see how I could get the top to factor out the top and cancel to get this fraction. I have a strong feeling that I did something wrong but do not know where or how to fix this.

    Thank you.
    Last edited by IDontunderstand; December 9th 2010 at 06:53 PM. Reason: show my work
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  2. #2
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    \displaystyle \frac{2\tan{x}}{1 - \tan^2{x}} + \frac{1}{2\cos^2{x} - 1} = \frac{\frac{2\sin{x}}{\cos{x}}}{1 - \frac{\sin^2{x}}{\cos^2{x}}} + \frac{1}{2\cos^2{x} - 1}

    \displaystyle = \frac{\frac{2\sin{x}}{\cos{x}}}{\frac{\cos^2{x} - \sin^2{x}}{\cos^2{x}}} + \frac{1}{2\cos^2{x} - 1}

    \displaystyle = \frac{2\sin{x}\cos{x}}{\cos^2{x} - \sin^2{x}} + \frac{1}{2\cos^2{x} - 1}

    \displaystyle = \frac{2\sin{x}\cos{x}}{\cos^2{x} - \sin^2{x}} + \frac{1}{\cos^2{x} + \cos^2{x} - 1}

    \displaystyle = \frac{2\sin{x}\cos{x}}{\cos^2{x} - \sin^2{x}} + \frac{1}{\cos^2{x} - \sin^2{x}}

    \displaystyle = \frac{2\sin{x}\cos{x} + 1}{\cos^2{x} - \sin^2{x}}

    \displaystyle = \frac{\cos^2{x} + 2\sin{x}\cos{x} + \sin^2{x}}{\cos^2{x} - \sin^2{x}}

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

    \displaystyle = \frac{\cos{x} + \sin{x}}{\cos{x} - \sin{x}}.
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  3. #3
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    Trig ident.

    Quote Originally Posted by IDontunderstand View Post
    I have gotten stuck again on another problem.

    [LaTeX ERROR: Convert failed]

    So this is what I did was made one fraction in the denominator:

    [LaTeX ERROR: Convert failed]

    Notice that the left hand side is:

    \displaystyle \tan 2x +\frac{1}{\cos (2x)}

    I think the right hand side might be more fruitful to work on.

    \displaystyle \frac{\cos x + \sin x}{\cos x - \sin x}

    Multiply top & bottom by \cos x + \sin x.

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

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

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

    That should help.
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