Showing $8(\arctan(\sqrt2-1)) = \pi$

Question

My friend is saying $$8(\arctan(\sqrt2-1)) = \pi$$ I know $4\arctan(1) = \pi$

I thought of $4\arctan(\sqrt2-1) + 4\arctan(\sqrt2-1)$

But I got stuck somewhere

How to solve?

Answer 1

So you have to prove $$\tan {\pi\over 8} = \sqrt{2}-1$$

Notice that $$1= \tan (2\cdot {\pi\over 8}) = {2\tan {\pi\over 8}\over 1-\tan ^2{\pi\over 8}}$$

and solve it on $\tan {\pi\over 8}$.

Answer 2

After taking the $ \tan $ on both sides you get: $$ \tan\frac{\pi}{8} = \sqrt{2}-1 $$ Note these two double angle formulas $ \sin2x = 2\sin x \cos x $ and $ \cos 2x = 2\cos^2x-1 $. Now we have: $$ \tan\frac{\pi}{8} = \frac{\sin\frac{\pi}{8}}{\cos\frac{\pi}{8}} = \frac{\sin\frac{\pi}{8}}{\cos\frac{\pi}{8}} \cdot \frac{\cos\frac{\pi}{8}}{\cos\frac{\pi}{8}} = \frac{\sin\frac{\pi}{4}}{1+\cos\frac{\pi}{4}} = \frac{\sqrt{2}/2}{1+\sqrt{2}/2} = \sqrt{2} - 1$$

Answer 3

Your friend is saying $$8(\arctan(\sqrt2-1)) = \pi$$ in other words removing inverse function he is saying $$ (\sqrt2-1)=\tan \frac{\pi}{8}=t $$ that you can say using double angle formula for tan $$\tan \pi/4 = \frac{2t}{1-t^2} $$ which is same thing as saying $$ 1=\frac{2 (\sqrt2-1)}{1-(3-2\sqrt2)}$$ thats same thing as saying ... after cancelling $2$ in numerator & denominator: $$ 1=\frac{\sqrt2-1}{\sqrt2-1}$$ So he is right.