I did it as follows: $$\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)=\tan\Bigg(\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)\Bigg)=\frac{\sqrt{2}-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}\sqrt{2}}=\frac{\sqrt{2}}{4}.$$ But there is no such an answer. What is wrong with it?
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4Where does the first equality come from?! – Wojowu Jun 30 '17 at 15:32
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What is the expected answer? Have you understood the method required? – lab bhattacharjee Jun 30 '17 at 15:34
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@Wojowu first, I have added tangent to the left-hand side of the problem and solved according to the formula $$\tan(x-y)=\frac{tan(x)-tan{y}}{1+tan{x}tan{y}}.$$ – user36339 Jun 30 '17 at 15:36
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1You can't write $x = \tan x$ and then expect to get the correct answer. What you found is the tangent of the number you seek, not the number itself. – Wojowu Jun 30 '17 at 15:38
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2Your very first equality is not true. Never use "=" this way. – Thomas Andrews Jun 30 '17 at 15:41
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1Because instead of finding $x$, he found $\tan(x)$ and then deduced what $x$ is. At no point he writes $x=\tan(x)$ or anything alike. – Wojowu Jun 30 '17 at 15:54
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The angle you are looking for is the red angle:
$$\arctan\sqrt{2}-\arctan\frac{1}{\sqrt{2}} = \arctan\left(\frac{\sqrt{2}-\frac{1}{\sqrt{2}}}{1+1}\right)=\color{red}{\arctan\frac{1}{\sqrt{8}}} $$
Jack D'Aurizio
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Could you briefly explain how you figured out that the graph looks like that? – user36339 Jul 01 '17 at 04:42
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1@Tug'tekin: by exploiting the definition of $\arctan$ and the fact that the diagonal of a unit square has length $\sqrt{2}$. – Jack D'Aurizio Jul 01 '17 at 06:03
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$$\tan[\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)]=\frac{\sqrt{2}-1/\sqrt{2}}{1+1}=\frac{\sqrt{2}}{4},$$
on simplifying we see that
$$\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right) =\arctan(\frac{\sqrt{2}}{4})= \sin^{-1} \frac{1}{3}.$$
BTW, this identity suggests a Ruler & Compass construction to trisect a line segment $OA$ (Somos' trig simplification, Jack D'Aurizio's construction are put together):
Narasimham
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I think your are right! Your solution is true!
Just $\frac{\sqrt2}{4}=\frac{(\sqrt2)^2}{4\sqrt2}=\frac{1}{2\sqrt2}=\frac{1}{\sqrt8}$
Michael Rozenberg
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I have substituted tangent to the left side just as you did it on https://math.stackexchange.com/questions/2339063/find-arcsin-frac45-arccos-frac1-sqrt50. But here, this method turned out to be wrong as Wojowu said above, could you briefly explain why? – user36339 Jul 01 '17 at 04:49
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1@ I can show, how we can do it with $\tan$. It works! Just we need to remember that $\frac{\pi}{2}<\arcsin\frac{4}{5}+\arccos\frac{1}{\sqrt{50}}<\pi$, but $-\frac{\pi}{2}<\arctan{a}<\frac{\pi}{2}$. – Michael Rozenberg Jul 01 '17 at 05:16