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About LinkBacksSimplify the expressionMy final answer is
but its telling me I am wrong
this then works down eventually to my answer or is my frist step wrong?
I found my mistake the answer is
First, I would writeOriginally Posted by M670
Simplify the expressionMy final answer is
I found my mistake the answer is(ignoring the plus/minus signs for the moment). Then that means
Then make use ofso that
![\displaystyle \begin{align*} \frac{\sqrt{1-\cos^2{\left[2\arccos{\left(\frac{x}{4}\right)}\right]}}}{\cos{\left[2\arccos{\left(\frac{x}{4}\right)}\right]}} &= \frac{\sqrt{1 - \left\{ 2\cos^2{\left[\arccos{\left(\frac{x}{4}\right)}\right]} - 1 \right\}^2}}{2\cos^2{\left[\arccos{\left(\frac{x}{4}\right)}\right]}-1} \\ &= \frac{\sqrt{1 - \left[ 2\left(\frac{x}{4}\right)^2 - 1 \right]^2}}{2\left(\frac{x}{4}\right)^2-1} \\ &= \frac{\sqrt{1 - \left[ 2\left( \frac{x^2}{16} \right) - 1 \right]^2}}{2\left( \frac{x^2}{16} \right) - 1} \\ &= \frac{\sqrt{1 - \left( \frac{x^2 - 8}{8} \right)^2}}{\frac{x^2 - 8}{8}} \\ &= \frac{8\sqrt{\frac{8^2 - \left( x^2 - 8 \right)^2}{8^2}}}{x^2 - 8} \\ &= \frac{\sqrt{8^2 - \left( x^2 - 8 \right)^2}}{x^2 - 8} \\ &= \frac{\sqrt{\left[ 8 - \left(x^2 - 8 \right) \right] \left[ 8 + \left( x^2 - 8 \right) \right] }}{x^2 - 8} \\ &= \frac{\sqrt{x^2 \left( 16 - x^2 \right) }}{x^2 - 8} \\ &= \frac{ x \sqrt{ 16 - x^2 } }{ x^2 - 8 } \end{align*}](http://latex.codecogs.com/png.latex?\displaystyle \begin{align*} \frac{\sqrt{1-\cos^2{\left[2\arccos{\left(\frac{x}{4}\right)}\right]}}}{\cos{\left[2\arccos{\left(\frac{x}{4}\right)}\right]}} &= \frac{\sqrt{1 - \left\{ 2\cos^2{\left[\arccos{\left(\frac{x}{4}\right)}\right]} - 1 \right\}^2}}{2\cos^2{\left[\arccos{\left(\frac{x}{4}\right)}\right]}-1} \\ &= \frac{\sqrt{1 - \left[ 2\left(\frac{x}{4}\right)^2 - 1 \right]^2}}{2\left(\frac{x}{4}\right)^2-1} \\ &= \frac{\sqrt{1 - \left[ 2\left( \frac{x^2}{16} \right) - 1 \right]^2}}{2\left( \frac{x^2}{16} \right) - 1} \\ &= \frac{\sqrt{1 - \left( \frac{x^2 - 8}{8} \right)^2}}{\frac{x^2 - 8}{8}} \\ &= \frac{8\sqrt{\frac{8^2 - \left( x^2 - 8 \right)^2}{8^2}}}{x^2 - 8} \\ &= \frac{\sqrt{8^2 - \left( x^2 - 8 \right)^2}}{x^2 - 8} \\ &= \frac{\sqrt{\left[ 8 - \left(x^2 - 8 \right) \right] \left[ 8 + \left( x^2 - 8 \right) \right] }}{x^2 - 8} \\ &= \frac{\sqrt{x^2 \left( 16 - x^2 \right) }}{x^2 - 8} \\ &= \frac{ x \sqrt{ 16 - x^2 } }{ x^2 - 8 } \end{align*})
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