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Pi/4=8arctan(1/10) - arctan(a/b). Find "a" and "b", so that the equality is true. Thank you very much for any help.

 Dec 2, 2015

Best Answer 

 #2
avatar+23278 
+40

Pi/4=8arctan(1/10) - arctan(a/b). Find "a" and "b", so that the equality is true. Thank you very much for any help.

 

\(\begin{array}{rcl} 8\cdot \arctan{ ( \frac{1}{10} ) } - \arctan{ ( \frac{a}{b} ) } &=& \frac{\pi}{4}\\ \end{array}\)

 

\( \text{Formula: } \quad \boxed{~ \tan{(2x)} = \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} } ~}\)

 

I.  We set:    \(x = \arctan{ ( \frac{1}{10} ) } \quad \text{ and } \quad \tan{(x)} = \frac{1}{10}\)

\(\small{ \begin{array}{rcl} \tan{(2x)} = \tan{( 2 \arctan{ ( \frac{1}{10} ) } )} &=& \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} }= \dfrac{2\cdot \frac{1}{10} } { 1- \frac{1}{10}\cdot \frac{1}{10} } = \frac{20}{99}\\ \qquad 2 \arctan{ ( \frac{1}{10} ) } &=& \arctan{ ( \frac{20}{99} ) } \end{array} }\)

 

 

II. We set:   \(x = 2\arctan{ ( \frac{1}{10} ) } = \arctan{ ( \frac{20}{99} ) }\quad \text{ and } \quad \tan{(x)} = \frac{20}{99} \)

\(\small{ \begin{array}{rcl} \tan{(2x)} = \tan{( 4 \arctan{ ( \frac{1}{10} ) } )} &=& \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} } =\dfrac{2\cdot \frac{20}{99} } { 1- \frac{20}{99}\cdot \frac{20}{99} } = \frac{3960}{9401}\\ \qquad 4 \arctan{ ( \frac{1}{10} ) } &=& \arctan{ ( \frac{3960}{9401} ) } \end{array} }\)

 

 

III. We set:   \(x = 4\arctan{ ( \frac{1}{10} ) } = \arctan{ ( \frac{3960}{9401} ) }\quad \text{ and } \quad \tan{(x)} = \frac{3960}{9401} \)

\(\small{ \begin{array}{rcl} \tan{(2x)} = \tan{( 8 \arctan{ ( \frac{1}{10} ) } )} &=& \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} } =\dfrac{2\cdot \frac{3960}{9401} } { 1- \frac{3960}{9401}\cdot \frac{3960}{9401} } = \frac{74455920}{72697201}\\ \qquad 8 \arctan{ ( \frac{1}{10} ) } &=& \arctan{ ( \frac{74455920}{72697201} ) } \end{array} }\)

 

 

IV.

\(\text{Formula: } \quad \boxed{~ \tan{(x+y)} = \frac{ \tan{(x)}+\tan{(y)} }{ 1- \tan{(x)}\cdot \tan{(y)} } ~}\)

 

We set:   \(x = 8\arctan{ ( \frac{1}{10} ) } = \arctan{ ( \frac{74455920}{72697201} ) } \quad \text{ and } \quad \tan{(x)} = \frac{74455920}{72697201}\\ y = \arctan{ ( \frac{a}{b} ) } \quad \text{ and } \quad \tan{(y)} = \frac{a}{b}\\ \tan{(x+y)} = 1\)

 

\(\small{ \begin{array}{rcl} \tan{(x+y)} = \frac{ \tan{(x)}+\tan{(y)} }{ 1- \tan{(x)}\cdot \tan{(y)} } &=& \dfrac{ \frac{74455920}{72697201}+\frac{a}{b} }{ 1- \frac{74455920}{72697201}\cdot \frac{a}{b} } = 1\\\\ \qquad \dfrac{ \frac{74455920}{72697201}+\frac{a}{b} }{ 1- \frac{74455920}{72697201}\cdot \frac{a}{b} } &=& 1\\\\ \qquad \frac{74455920}{72697201}+\frac{a}{b} &=& 1- \frac{74455920}{72697201}\cdot \frac{a}{b}\\\\ \qquad \frac{a}{b}+ \frac{74455920}{72697201}\cdot \frac{a}{b} &=& 1-\frac{74455920}{72697201}\\\\ \qquad \frac{a}{b}\cdot \left( 1 + \frac{74455920}{72697201} \right) &=& 1-\frac{74455920}{72697201}\\\\ \qquad \frac{a}{b}\cdot \left( \frac{72697201+74455920}{72697201} \right) &=& \frac{72697201-74455920}{72697201}\\\\ \qquad \frac{a}{b}\cdot ( 72697201+74455920 ) &=& 72697201-74455920\\\\ \qquad \frac{a}{b} &=& \dfrac{ 72697201-74455920 } { 72697201+74455920 }\\\\ \qquad \frac{a}{b} &=& \dfrac{ -1758719 } { 147153121 }\\\\ \hline \\ \tan{(x+y)} = 1 \\ \tan{(8\arctan{ ( \frac{1}{10} ) }+\arctan{ ( \frac{a}{b} ) })} &=& 1 \\ \tan{(8\arctan{ ( \frac{1}{10} ) }-\arctan{ ( \frac{ 1758719 } { 147153121 } ) })} &=& 1 \\ 8\arctan{ ( \frac{1}{10} ) }-\arctan{ ( \frac{ 1758719 } { 147153121 } ) } &=& \arctan{(1)} \\\\ \mathbf{ 8\arctan{ ( \dfrac{1}{10} ) }-\arctan{ ( \dfrac{ 1758719 } { 147153121 } ) } } & \mathbf{=} & \mathbf{\dfrac{\pi} {4}} \end{array} }\)

 

laugh

 Dec 2, 2015
 #2
avatar+23278 
+40
Best Answer

Pi/4=8arctan(1/10) - arctan(a/b). Find "a" and "b", so that the equality is true. Thank you very much for any help.

 

\(\begin{array}{rcl} 8\cdot \arctan{ ( \frac{1}{10} ) } - \arctan{ ( \frac{a}{b} ) } &=& \frac{\pi}{4}\\ \end{array}\)

 

\( \text{Formula: } \quad \boxed{~ \tan{(2x)} = \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} } ~}\)

 

I.  We set:    \(x = \arctan{ ( \frac{1}{10} ) } \quad \text{ and } \quad \tan{(x)} = \frac{1}{10}\)

\(\small{ \begin{array}{rcl} \tan{(2x)} = \tan{( 2 \arctan{ ( \frac{1}{10} ) } )} &=& \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} }= \dfrac{2\cdot \frac{1}{10} } { 1- \frac{1}{10}\cdot \frac{1}{10} } = \frac{20}{99}\\ \qquad 2 \arctan{ ( \frac{1}{10} ) } &=& \arctan{ ( \frac{20}{99} ) } \end{array} }\)

 

 

II. We set:   \(x = 2\arctan{ ( \frac{1}{10} ) } = \arctan{ ( \frac{20}{99} ) }\quad \text{ and } \quad \tan{(x)} = \frac{20}{99} \)

\(\small{ \begin{array}{rcl} \tan{(2x)} = \tan{( 4 \arctan{ ( \frac{1}{10} ) } )} &=& \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} } =\dfrac{2\cdot \frac{20}{99} } { 1- \frac{20}{99}\cdot \frac{20}{99} } = \frac{3960}{9401}\\ \qquad 4 \arctan{ ( \frac{1}{10} ) } &=& \arctan{ ( \frac{3960}{9401} ) } \end{array} }\)

 

 

III. We set:   \(x = 4\arctan{ ( \frac{1}{10} ) } = \arctan{ ( \frac{3960}{9401} ) }\quad \text{ and } \quad \tan{(x)} = \frac{3960}{9401} \)

\(\small{ \begin{array}{rcl} \tan{(2x)} = \tan{( 8 \arctan{ ( \frac{1}{10} ) } )} &=& \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} } =\dfrac{2\cdot \frac{3960}{9401} } { 1- \frac{3960}{9401}\cdot \frac{3960}{9401} } = \frac{74455920}{72697201}\\ \qquad 8 \arctan{ ( \frac{1}{10} ) } &=& \arctan{ ( \frac{74455920}{72697201} ) } \end{array} }\)

 

 

IV.

\(\text{Formula: } \quad \boxed{~ \tan{(x+y)} = \frac{ \tan{(x)}+\tan{(y)} }{ 1- \tan{(x)}\cdot \tan{(y)} } ~}\)

 

We set:   \(x = 8\arctan{ ( \frac{1}{10} ) } = \arctan{ ( \frac{74455920}{72697201} ) } \quad \text{ and } \quad \tan{(x)} = \frac{74455920}{72697201}\\ y = \arctan{ ( \frac{a}{b} ) } \quad \text{ and } \quad \tan{(y)} = \frac{a}{b}\\ \tan{(x+y)} = 1\)

 

\(\small{ \begin{array}{rcl} \tan{(x+y)} = \frac{ \tan{(x)}+\tan{(y)} }{ 1- \tan{(x)}\cdot \tan{(y)} } &=& \dfrac{ \frac{74455920}{72697201}+\frac{a}{b} }{ 1- \frac{74455920}{72697201}\cdot \frac{a}{b} } = 1\\\\ \qquad \dfrac{ \frac{74455920}{72697201}+\frac{a}{b} }{ 1- \frac{74455920}{72697201}\cdot \frac{a}{b} } &=& 1\\\\ \qquad \frac{74455920}{72697201}+\frac{a}{b} &=& 1- \frac{74455920}{72697201}\cdot \frac{a}{b}\\\\ \qquad \frac{a}{b}+ \frac{74455920}{72697201}\cdot \frac{a}{b} &=& 1-\frac{74455920}{72697201}\\\\ \qquad \frac{a}{b}\cdot \left( 1 + \frac{74455920}{72697201} \right) &=& 1-\frac{74455920}{72697201}\\\\ \qquad \frac{a}{b}\cdot \left( \frac{72697201+74455920}{72697201} \right) &=& \frac{72697201-74455920}{72697201}\\\\ \qquad \frac{a}{b}\cdot ( 72697201+74455920 ) &=& 72697201-74455920\\\\ \qquad \frac{a}{b} &=& \dfrac{ 72697201-74455920 } { 72697201+74455920 }\\\\ \qquad \frac{a}{b} &=& \dfrac{ -1758719 } { 147153121 }\\\\ \hline \\ \tan{(x+y)} = 1 \\ \tan{(8\arctan{ ( \frac{1}{10} ) }+\arctan{ ( \frac{a}{b} ) })} &=& 1 \\ \tan{(8\arctan{ ( \frac{1}{10} ) }-\arctan{ ( \frac{ 1758719 } { 147153121 } ) })} &=& 1 \\ 8\arctan{ ( \frac{1}{10} ) }-\arctan{ ( \frac{ 1758719 } { 147153121 } ) } &=& \arctan{(1)} \\\\ \mathbf{ 8\arctan{ ( \dfrac{1}{10} ) }-\arctan{ ( \dfrac{ 1758719 } { 147153121 } ) } } & \mathbf{=} & \mathbf{\dfrac{\pi} {4}} \end{array} }\)

 

laugh

heureka Dec 2, 2015
 #3
avatar+105509 
+5

Thanks Heureka, that is a fabulous answer. :)

 Dec 2, 2015
 #4
avatar
+10

Brilliant work, heureka!. Thank you very much for  the outstanding technical work and the final answer. I noticed that both "a" and "b" are primes! Wasn't expecting that at all. A brief question: couldn't this be solved by " continued fraction" method? That is: Pi/4= 8arctan(1/10) - arctan(1)=0.011951......etc. Then we take the tangent of this, which comes to=0.011951625545203353.......which is the answer of your two numbers: 1,758,719 / 147,153,121=0.011951625545203353........etc.

 

P.S. I plugged this into "WolframAlpha" and sure enough it gives this continued fraction:

[0; 83, 1, 2, 27, 1, 2, 1, 1, 1, 2, 3, 4, 6, 1, 6], When you work this backwards, it gives the fraction you found!:1758719/147153121=0.01195162554520335317930497716049121377452810.....

THANKS A LOT.

 Dec 2, 2015

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