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Math Help - Need help simplifying this trigonometric substitution problem on homework.

  1. #1
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    Need help simplifying this trigonometric substitution problem on homework.

    I need help simplifying (sqrt(x^2-49))/(x) where x =7sec(theta)

    I'm completely stumped, Trigonometry has proven to not be my strongest subject. Any help, or something worked out to push me in the right direction, would be fantastic!


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  2. #2
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    Re: Need help simplifying this trigonometric substitution problem on homework.

    sqrt(x^2 - 49)/x where x = 7 sec(theta)
    sqrt(7(sec(theta)))^2 - 49) / (7sec(theta))
    sqrt(49(sec(theta))^2 - 49)/ (7sec(theta))
    sqrt(49[(sec(theta))^2 - 1]) / (7sec(theta))
    sqrt{49(tan(theta))^2 / (7sec(theta))
    (7 tan(theta)) / (7sec(theta))
    7 tan(theta) cos(theta)/ 7
    tan(theta) cos(theta)
    sin(theta)/cos(theta) * cos(theta)
    sin(theta)
    Last edited by mathguy25; April 10th 2013 at 04:54 PM.
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  3. #3
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    Re: Need help simplifying this trigonometric substitution problem on homework.

    Hello, AKPunk49!

    I'll write it out in LaTeX . . . may be easier to read.


    \text{Simplify: }\:\frac{\sqrt{x^2-49}}{x},\:\text{ where }x = 7\sec\theta

    \frac{\sqrt{x^2-49}}{x} \;=\;\frac{\sqrt{(7\sec\theta)^2 - 49}}{7\sec\theta} \;=\;\frac{\sqrt{49\sec^2\!\theta - 49}}{7\sec\theta}

    . . . . . . . =\;\frac{\sqrt{49(\sec^2\!\theta-1)}}{\7\sec\theta} \;=\; \frac{\sqrt{49\tan^2\!\theta}}{7\sec\theta}

    . . . . . . . =\;\frac{7\tan\theta}{7\sec\theta} \;=\;\frac{\tan\theta}{\sec\theta} \;=\;\frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}  {\cos\theta}} \;=\;\sin\theta
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