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what is x if the equation is 2(tan-1(x/160)) = tan-1(x/60)?

Guest Jul 9, 2015

Best Answer 

 #1
avatar+18712 
+10

what is x if the equation is 2(tan-1(x/160)) = tan-1(x/60) ?

 

$$\small{\text{$
\begin{array}{rclrcl}
2 \cdot \left[ \arctan{ \left( \dfrac{ x } { 160 } \right) } \right]
&=& \arctan{ \left( \dfrac{ x } { 60 } \right) } \qquad | \qquad
\varphi = \arctan{ \left( \dfrac{ x } { 160 } \right) } \\\\
2\cdot \varphi &=& \arctan{ \left( \dfrac{ x } { 60 } \right) }
\qquad |\qquad \tan{()}\\\\
\tan{ (2\cdot \varphi) } &=& \dfrac{x}{60} \\\\
&& &\tan{ (2\cdot \varphi) }
&=& \dfrac{ 2\cdot\tan{(\varphi)} } { 1-[\tan{(\varphi)]^2} } \\\\\
&& &\tan{ (2\cdot \varphi) }
&=& \dfrac{ 2\cdot\tan{(
\arctan{ \left( \dfrac{ x } { 160 } \right) }
)} } { 1-[\tan{(
\arctan{ \left( \dfrac{ x } { 160 } \right) }
)]^2} } \\\\\
&& &\tan{ (2\cdot \varphi) }
&=& \dfrac{ 2\cdot \dfrac{ x } { 160 } }
{ 1- \left( \dfrac{ x } { 160 } \right)^2 } \\\\\
\dfrac{ 2\cdot \dfrac{ x } { 160 } }
{ 1- \left( \dfrac{ x } { 160 } \right)^2 } &=& \dfrac{x}{60} \\\\
\dfrac{ 1 }
{ 1- \left( \dfrac{ x } { 160 } \right)^2 } &=& \dfrac{4}{3} \\\\
1- \left( \dfrac{ x } { 160 } \right)^2 &=& \dfrac{3}{4} \\\\
\left( \dfrac{ x } { 160 } \right)^2 &=& \dfrac{1}{4} \qquad | \qquad \pm\sqrt{}\\\\
\dfrac{ x } { 160 } &=& \pm0.5 \\\\
x &=& \pm0.5 \cdot 160 \\\\
\mathbf{x_1} & \mathbf{=} & \mathbf{80} \\\\
\mathbf{x_2} & \mathbf{=} & \mathbf{-80} \\\\
\end{array}
$}}$$

 

heureka  Jul 10, 2015
Sort: 

3+0 Answers

 #1
avatar+18712 
+10
Best Answer

what is x if the equation is 2(tan-1(x/160)) = tan-1(x/60) ?

 

$$\small{\text{$
\begin{array}{rclrcl}
2 \cdot \left[ \arctan{ \left( \dfrac{ x } { 160 } \right) } \right]
&=& \arctan{ \left( \dfrac{ x } { 60 } \right) } \qquad | \qquad
\varphi = \arctan{ \left( \dfrac{ x } { 160 } \right) } \\\\
2\cdot \varphi &=& \arctan{ \left( \dfrac{ x } { 60 } \right) }
\qquad |\qquad \tan{()}\\\\
\tan{ (2\cdot \varphi) } &=& \dfrac{x}{60} \\\\
&& &\tan{ (2\cdot \varphi) }
&=& \dfrac{ 2\cdot\tan{(\varphi)} } { 1-[\tan{(\varphi)]^2} } \\\\\
&& &\tan{ (2\cdot \varphi) }
&=& \dfrac{ 2\cdot\tan{(
\arctan{ \left( \dfrac{ x } { 160 } \right) }
)} } { 1-[\tan{(
\arctan{ \left( \dfrac{ x } { 160 } \right) }
)]^2} } \\\\\
&& &\tan{ (2\cdot \varphi) }
&=& \dfrac{ 2\cdot \dfrac{ x } { 160 } }
{ 1- \left( \dfrac{ x } { 160 } \right)^2 } \\\\\
\dfrac{ 2\cdot \dfrac{ x } { 160 } }
{ 1- \left( \dfrac{ x } { 160 } \right)^2 } &=& \dfrac{x}{60} \\\\
\dfrac{ 1 }
{ 1- \left( \dfrac{ x } { 160 } \right)^2 } &=& \dfrac{4}{3} \\\\
1- \left( \dfrac{ x } { 160 } \right)^2 &=& \dfrac{3}{4} \\\\
\left( \dfrac{ x } { 160 } \right)^2 &=& \dfrac{1}{4} \qquad | \qquad \pm\sqrt{}\\\\
\dfrac{ x } { 160 } &=& \pm0.5 \\\\
x &=& \pm0.5 \cdot 160 \\\\
\mathbf{x_1} & \mathbf{=} & \mathbf{80} \\\\
\mathbf{x_2} & \mathbf{=} & \mathbf{-80} \\\\
\end{array}
$}}$$

 

heureka  Jul 10, 2015
 #2
avatar+26326 
+5

There is also the trivial solution x= 0

.

Alan  Jul 10, 2015
 #3
avatar+90996 
+5

I have only just now had a chance to look at your answer Heureka.

I really like it.      

 

 

I have added this thread address to our "Great Answers to Learn From" sticky thread.  :)

Melody  Jul 16, 2015

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