# Where have I made a mistake when comparing $2\arctan(2\sqrt{2}-1)$ and $3\arctan\left(\frac{1}{4}\right)+\arctan\left(\frac{5}{99}\right)$?

Withough using a calculator, find which of $$2\arctan(2\sqrt{2}-1)$$ or $$3\arctan\left(\frac{1}{4}\right)+\arctan\left(\frac{5}{99}\right)$$ is bigger.

I used the formula $$\arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy}\right)$$ to solve this question. Right hand side I found $$3\arctan\left(\frac{1}{4}\right)+\arctan\left(\frac{5}{99}\right)$$ equals to $$\arctan(1)$$ and it matches with calculator solution. For left hand side it's not that good. I found $$2\arctan(2\sqrt{2}-1)=\arctan\left(-\frac{2+3\sqrt{2}}{4}\right)$$ and I said $$3\arctan\left(\frac{1}{4}\right)+\arctan\left(\frac{5}{99}\right)$$ is the bigger one because of the graph of $$\arctan(x)$$. But calculator solution is not matches with my solution. Where am I making a mistake. Thanks for all help!

Hint:

Your mistake is hidden in the fact that$$\frac\pi2<\phi\equiv 2\arctan (2\sqrt2-1)<\pi,$$ the negative sign of $$\tan(\phi)$$ being a clear alarm bell.

If you ignore the bell you will obtain $$\arctan(\tan\phi)=\phi-\pi,$$ since the range of $$\arctan x$$ is $$\left[-\frac\pi2,\frac\pi2\right]$$.

• Thank you so much i realized my mistake and fixed my solution! – Topological Jun 11 at 17:22

The arctangent addition formula must be interpreted carefully, because it doesn't hold as written for all $$x$$ and $$y$$. In particular it cannot hold if $$|\arctan(x)+\arctan(y)|\gt\pi/2$$, because the range of the arctangent function is $$(-\pi/2,\pi/2)$$. That seems to be the proximate cause of your mistake. (The full, correct addition formula can found in the link in lab bhattacharjee's comment below the OP.)

In this case, it's relatively easy to see that

$$2\arctan(2\sqrt2-1)\gt2\arctan1=\pi/2\gt1\gt3/4+5/99\gt3\arctan(1/4)+\arctan(5/99)$$

using just the increasing nature of the arctangent function for the first inequality and the inequality $$x\gt\arctan x$$ for $$x\gt0$$ for the final inequality. The inequality $$x\gt\arctan x$$ for $$x\gt0$$ can be seen from

$$\arctan x=\int_0^x{dt\over1+t^2}\lt\int_0^x{dt\over1+0}=x$$

Remark: It's possible the intended comparison was actually between $$3\arctan(1/4)+\arctan(5/99)$$ and $$2\arctan(\sqrt2-1)$$, not $$2\arctan(2\sqrt2-1)$$.