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Math Help - Trig Proof, help please!

  1. #1
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    Trig Proof, help please!

    secx = (2(cosxsin2x - sinxcos2x))/sin2x
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  2. #2
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    Quote Originally Posted by sophiaroth View Post
    secx = (2(cosxsin2x - sinxcos2x))/sin2x
    \frac{2[\cos{x}\sin(2x) - \sin{x}\cos(2x)]}{\sin(2x)}

    \frac{2[\cos{x} \cdot 2\sin{x}\cos{x} - \sin{x}(2\cos^2{x}-1)]}{2\sin{x}\cos{x}}

    \frac{2[2\cos^2{x}\sin{x} - 2\cos^2{x}\sin{x} + \sin{x}]}{2\sin{x}\cos{x}}

    \frac{2\sin{x}}{2\sin{x}\cos{x}}

    \frac{1}{\cos{x}} = \sec{x}
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  3. #3
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    thanks! i had tried cos2x = cos^2x - sin^2x, and never went as far as 2cos^2x - 1
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  4. #4
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    Quote Originally Posted by sophiaroth View Post
    secx = (2(cosxsin2x - sinxcos2x))/sin2x
    \text{this is sin(2x-x)}
    \frac{2\overbrace{(\cos{x}\sin(2x) - \sin{x}\cos(2x))}}{\sin(2x)}

    =\frac{2\sin{x}}{2\sin{x}\cos{x}}

    =\frac{1}{\cos{x}} = \sec{x}

    just to get rid of few steps
    Last edited by great_math; June 19th 2009 at 03:20 AM.
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