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tan(2cos^−1(x/3)) simplified?

Guest Feb 27, 2017
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simplify tan[ 2*arccos(x/3) ]

 

\(\begin{array}{|rcll|} \hline \cos(\varphi) &=& \frac{x}{3} \\ \varphi &=& \arccos\left(\frac{x}{3} \right) \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \tan(2 \varphi ) &=& \frac{\sin(2 \varphi)}{\cos(2 \varphi)} \quad & | \quad \sin(2 \varphi) = 2 \sin(\varphi)\cos(\varphi) \quad \cos(2\ \varphi) = 2 \cos^2(\varphi)-1 \\ \tan(2 \varphi ) &=& \frac{2 \sin(\varphi)\cos(\varphi)}{2 \cos^2(\varphi)-1} \quad & | \quad \cos(\varphi) = \frac{x}{3} \\ \tan(2 \varphi ) &=& \frac{2 \sin(\varphi)\frac{x}{3}}{2 (\frac{x}{3})^2-1} \quad & | \quad \sin(\varphi) = \sin(\arccos\left(\frac{x}{3} \right)) \\ \tan(2 \varphi ) &=& \frac{2 \sin(\arccos\left(\frac{x}{3} \right))\frac{x}{3}}{2 (\frac{x}{3})^2-1} \\ \hline \end{array} \)

 

 

\(\begin{array}{|rcll|} \hline \tan(2 \varphi ) &=& \frac{2 \sin(\arccos\left(\frac{x}{3} \right))\frac{x}{3}}{2 (\frac{x}{3})^2-1} \quad & | \quad \sin(\arccos\left(\frac{x}{3} \right)) = \sqrt{1-\left(\dfrac{x}{3}\right)^2} \\ \tan(2 \varphi ) &=& \frac{2 \sqrt{1-\left(\frac{x}{3}\right)^2}\frac{x}{3}}{2 (\frac{x}{3})^2-1} \\ \tan(2 \varphi ) &=& \frac{2 \sqrt{\frac{9-x^2}{9}}\frac{x}{3}}{ \frac{2x^2-9}{9}} \\ \tan(2 \varphi ) &=& \frac{2 \frac{\sqrt{9-x^2}}{3}\frac{x}{3}}{ \frac{2x^2-9}{9}} \\ \tan(2 \varphi ) &=& \frac{2 \frac{\sqrt{9-x^2}}{9}x}{ \frac{2x^2-9}{9}} \\ \tan(2 \varphi ) &=& \frac{2 \sqrt{9-x^2} x}{ 2x^2-9 } \quad & | \quad \sqrt{9-x^2} = \sqrt{3-x}\cdot\sqrt{3+x}\\ \tan(2 \varphi ) &=& \frac{2 \sqrt{3-x}\cdot x \cdot \sqrt{3+x}}{ 2x^2-9 } \\ \mathbf{\tan(2 \arccos\left(\frac{x}{3} \right) ) }& \mathbf{=} & \mathbf{\frac{2 \sqrt{3-x}\cdot x \cdot \sqrt{3+x}}{ 2x^2-9 } } \\ \hline \end{array}\)

 

laugh

heureka  Feb 27, 2017

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