Simplify

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November 24th 2012, 07:33 AM
M670

Simplify

Simplify the expression
http://webwork.mathstat.concordia.ca...3958a968b1.png

My final answer is $\frac{2\sqrt{16-x^2}}{2x^2-16}$ but its telling me I am wrong

$\frac{2\tan(\alpha)}{1-\tan^2(\alpha)}$ this then works down eventually to my answer or is my frist step wrong?

I found my mistake the answer is $\frac{2x\sqrt{16-x^2}}{2x^2-16}$
November 24th 2012, 07:57 AM
Soroban

Re: Simplify

Hello, M670!

You are missing an $x.$

Quote:

$\text{Simplify: }\;\tan\left[2\cos^{-1}\left(\frac{x}{4}\right)\right]$

$\text{Let }\theta = \cos^{-1}\left(\frac{x}{4}\right) \quad\Rightarrow\quad \cos\theta \:=\:\frac{x}{4} \:=\:\frac{adj}{hyp}$

$\theta\text{ is in a right triangle with: }\,adj = x,\;hyp = 4$

$\text{Pythagorus says: }\,opp = \sqrt{16-x^2}$

$\text{Hence: }\:\tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{\sqrt{16-x^2}}{x}$

$\text{The problem becomes: }\:\tan2\theta \;=\;\dfrac{2\tan\theta}{1 - \tan^2\!\theta} \;=\;\dfrac{2\frac{\sqrt{16-x^2}}{x}}{1 - \frac{16-x^2}{x^2}}$

$\text{Multiply by }\frac{x^2}{x^2}\!:\;\;\dfrac{2x\sqrt{16-x^2}}{2x^2-16} \;=\;\frac{x\sqrt{16-x^2}}{x^2-8}$

You found your error . . . Good!
November 24th 2012, 05:32 PM
Prove It

Re: Simplify

Quote:

Originally Posted by M670

Simplify the expression
http://webwork.mathstat.concordia.ca...3958a968b1.png

My final answer is $\frac{2\sqrt{16-x^2}}{2x^2-16}$ but its telling me I am wrong

$\frac{2\tan(\alpha)}{1-\tan^2(\alpha)}$ this then works down eventually to my answer or is my frist step wrong?

I found my mistake the answer is $\frac{2x\sqrt{16-x^2}}{2x^2-16}$

First, I would write $\displaystyle \begin{align*} \tan{(\theta)} = \frac{\sin{(\theta)}}{\cos{(\theta)}} = \frac{\sqrt{1 - \cos^2{(\theta)}}}{\cos{(\theta)}} \end{align*}$ (ignoring the plus/minus signs for the moment). Then that means

$\displaystyle \begin{align*} \tan{\left[ 2\arccos{\left( \frac{x}{4} \right)} \right]} &= \frac{\sqrt{1 - \cos^2{\left[ 2\arccos{\left( \frac{x}{4} \right)} \right]}}}{\cos{\left[ 2\arccos{\left(\frac{x}{4}\right)}\right]}} \end{align*}$

Then make use of $\displaystyle \begin{align*} \cos{(2\theta)} = 2\cos^2{(\theta)} - 1 \end{align*}$ so that

$\displaystyle \begin{align*} \frac{\sqrt{1-\cos^2{\left[2\arccos{\left(\frac{x}{4}\right)}\right]}}}{\cos{\left[2\arccos{\left(\frac{x}{4}\right)}\right]}} &= \frac{\sqrt{1 - \left\{ 2\cos^2{\left[\arccos{\left(\frac{x}{4}\right)}\right]} - 1 \right\}^2}}{2\cos^2{\left[\arccos{\left(\frac{x}{4}\right)}\right]}-1} \\ &= \frac{\sqrt{1 - \left[ 2\left(\frac{x}{4}\right)^2 - 1 \right]^2}}{2\left(\frac{x}{4}\right)^2-1} \\ &= \frac{\sqrt{1 - \left[ 2\left( \frac{x^2}{16} \right) - 1 \right]^2}}{2\left( \frac{x^2}{16} \right) - 1} \\ &= \frac{\sqrt{1 - \left( \frac{x^2 - 8}{8} \right)^2}}{\frac{x^2 - 8}{8}} \\ &= \frac{8\sqrt{\frac{8^2 - \left( x^2 - 8 \right)^2}{8^2}}}{x^2 - 8} \\ &= \frac{\sqrt{8^2 - \left( x^2 - 8 \right)^2}}{x^2 - 8} \\ &= \frac{\sqrt{\left[ 8 - \left(x^2 - 8 \right) \right] \left[ 8 + \left( x^2 - 8 \right) \right] }}{x^2 - 8} \\ &= \frac{\sqrt{x^2 \left( 16 - x^2 \right) }}{x^2 - 8} \\ &= \frac{ x \sqrt{ 16 - x^2 } }{ x^2 - 8 } \end{align*}$