+0

# Find rational fraction

0
407
4

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.

Guest Dec 2, 2015

#2
+18956
+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:   $$x = 8\arctan{ ( \frac{1}{10} ) } = \arctan{ ( \frac{74455920}{72697201} ) } \quad \text{ and } \quad \tan{(x)} = \frac{74455920}{72697201}\\ y = \arctan{ ( \frac{a}{b} ) } \quad \text{ and } \quad \tan{(y)} = \frac{a}{b}\\ \tan{(x+y)} = 1$$

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

heureka  Dec 2, 2015
Sort:

#2
+18956
+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:   $$x = 8\arctan{ ( \frac{1}{10} ) } = \arctan{ ( \frac{74455920}{72697201} ) } \quad \text{ and } \quad \tan{(x)} = \frac{74455920}{72697201}\\ y = \arctan{ ( \frac{a}{b} ) } \quad \text{ and } \quad \tan{(y)} = \frac{a}{b}\\ \tan{(x+y)} = 1$$

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

heureka  Dec 2, 2015
#3
+91786
+5

Thanks Heureka, that is a fabulous answer. :)

Melody  Dec 2, 2015
#4
+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.

Guest Dec 2, 2015

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