Firstly used this formula $$ \begin{align} \arctan(\alpha)+\arctan(\beta) & =\arctan(\frac{1-xy}{x+y}),\quad x\gt0,y\gt0 \\ &=\arctan(\frac{1-\frac{1}{3}\frac{1}{9}}{\frac{1}{3}+\frac{1}{9}}) \\ &=\arctan(2) \end{align}$$ So it is $\arctan(2)+\arctan(\frac{7}{19}).$ Here I don't know what is the next step to solve it completely. A SIDE NOTE: AN EDIT HAS BEEN MADE TO THIS POST, I HAVE FOUND MY MISTAKE! NOW IT IS CLEAR TO ME, THANKS!
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3What is $(3+i)(9+i)(19+7i)$? – Angina Seng Jun 26 '17 at 18:43
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Perhaps this page would interest you: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/simpleTrig.html – Franklin Pezzuti Dyer Jun 26 '17 at 18:43
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Your formula is incorrect - you're missing a $\arctan$ term. – Nathaniel Bubis Jun 26 '17 at 18:44
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@lordsharktheunknown It is $7i^3+109i^2+931i.$ I would like to know what is the reason behind asking this question involving complex numbers. – user36339 Jun 26 '17 at 18:53
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@Tug'tekin I am sceptical about your numbers. You are aware though, that complex numbers have arguments? – Angina Seng Jun 26 '17 at 18:54
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@LordSharktheUnknown Actually, no. Because I haven't begun studying them yet. – user36339 Jun 26 '17 at 18:57
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@Tug'tekin Then you have lots of fun to look forward to! – Angina Seng Jun 26 '17 at 18:58
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@Tug'tekin, https://math.stackexchange.com/questions/1837410/inverse-trigonometric-function-identity-doubt-tan-1x-tan-1y-pi-tan – lab bhattacharjee Jun 27 '17 at 08:16
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The correct formula is: $$\arctan(u)+\arctan(v)=\arctan \left({\frac {u+v}{1-uv}}\right)$$
So: $$\arctan(1/3)+\arctan(1/9)=\arctan \left({\frac {6}{13}}\right)$$ $$\arctan(1/3)+\arctan(1/9)+\arctan(7/19)= \arctan \left({\frac {6/13+7/19}{1-6/13\cdot 7/19}}\right)$$ $$=\arctan(1)=\frac{\pi}{4}$$
Nathaniel Bubis
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Just calculate: $$\tan\left(\arctan\frac{1}{3}+\arctan\frac{1}{9}+\arctan\frac{7}{19}\right)=$$ $$=\frac{\frac{\frac{1}{3}+\frac{1}{9}}{1-\frac{1}{3}\cdot\frac{1}{9}}+\frac{7}{19}}{1-\frac{\frac{1}{3}+\frac{1}{9}}{1-\frac{1}{3}\cdot\frac{1}{9}}\cdot\frac{7}{19}}=1,$$ which gives the answer: $45^{\circ}$.
Michael Rozenberg
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