Find $\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)$

Question

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?

Answer 1

The angle you are looking for is the red angle:

$$\arctan\sqrt{2}-\arctan\frac{1}{\sqrt{2}} = \arctan\left(\frac{\sqrt{2}-\frac{1}{\sqrt{2}}}{1+1}\right)=\color{red}{\arctan\frac{1}{\sqrt{8}}} $$

Answer 2

$$\tan[\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)]=\frac{\sqrt{2}-1/\sqrt{2}}{1+1}=\frac{\sqrt{2}}{4},$$

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):

Answer 3

I think your are right! Your solution is true!

Just $\frac{\sqrt2}{4}=\frac{(\sqrt2)^2}{4\sqrt2}=\frac{1}{2\sqrt2}=\frac{1}{\sqrt8}$