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Math Help - trigonometry simplify

  1. #1
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    trigonometry simplify

    750/ \pi\*48^2 = ((sin (1+x))^2 - (sin 1)^2)/2

    Find the value of x.

    Please help, its making me cry!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by iphysics View Post
    750/ \pi\*48^2 = ((sin (1+x))^2 - (sin 1)^2)/2

    Find the value of x.

    Please help, its making me cry!
    Well, sin(1) is just a constant, so
    \frac{750}{\pi} \cdot 48^2 = \frac{sin^2(1 + x) - sin^2(1)}{2}

    2 \cdot 48^2 \cdot \frac{750}{\pi} = sin^2(1 + x) - sin^2(1)

    sin^2(1 + x) = 2 \cdot 48^2 \cdot \frac{750}{\pi} + sin^2(1)

    sin(1 + x) = \pm \sqrt{2 \cdot 48^2 \cdot \frac{750}{\pi} + sin^2(1)}

    x + 1 = sin^{-1} \left ( \pm \sqrt{2 \cdot 48^2 \cdot \frac{750}{\pi} + sin^2(1)} \right )

    x = -1 + sin^{-1} \left ( \pm \sqrt{2 \cdot 48^2 \cdot \frac{750}{\pi} + sin^2(1)} \right )

    Unfortunately 2 \cdot 48^2 \cdot 750/ \pi + sin^2(1) is rather larger than 1 so the value of x is imaginary, which is a result I'm assuming you don't want. So there is no real solution to this problem.

    -Dan
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by iphysics View Post
    750/ \pi\*48^2 = ((sin (1+x))^2 - (sin 1)^2)/2

    Find the value of x.

    Please help, its making me cry!
     <br />
[\sin(1+x)]^2=\frac{1500}{48^2 \pi}+ \sin(1)<br />

    so:

     <br />
\sin(1+x)=\sqrt{\frac{1500}{48^2 \pi}+ \sin(1)}<br />

    Then:

     <br />
x=\arcsin \left( \sqrt{\frac{1500}{48^2 \pi}+ \sin(1)}\right)-1<br />

    RonL
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by iphysics View Post
    750/ \pi\*48^2 = ((sin (1+x))^2 - (sin 1)^2)/2
    Quote Originally Posted by CaptainBlack View Post
     <br />
[\sin(1+x)]^2=\frac{1500}{48^2 \pi}+ \sin(1)<br />
    Well that makes more sense.

    @iphysics: PLEASE use parenthesis!

    -Dan
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