Trigonometric Arithmetic I

We have:

$\begin{cases} x+y=\tan^{-1}{ \frac { 2a }{1-a^2} }\\ \tan{x}\tan{y}=a^2,\end{cases}$ $\left| a \right| <1$

From the first equation, we find

$\tan{(x+y)}=\frac{2a}{1-a^2}$ ,

i.e.

$\frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}=\frac{2a}{1-a^2}$ .

Taking the second equation into account, we find

$\frac{\tan{x}+\tan{y}}{1-a^2}=\frac{2a}{1-a^2}$ ,

or

$\tan{x}+\tan{y}=2a$ .

From the system of equations

$\begin{cases} \tan{x}+\tan{y}=2a\\ \tan{x}\tan{y}=a^2\end{cases}$

we find $\tan{x}=a;$ $\tan{y}=a$ . From here, we see that

$x=180°n+\tan^{-1}{a},$ $y=180°m+\tan^{-1}a$ , $n\in {Z}$ .

But only one of these is arbitrary, since, in agreement with the first equation, the magnitude of $x+y$ , as the main function of the arctangent, must be bound within $-90°$ and $90°$ .

To deduce valid values of $n$ and $m$ , we substitute the found expressions into the first equation. We obtain

$180°(n+m)+2\tan^{-1}{a}=\tan^{-1}{\frac{2a}{1-a^2}}$ . $(A)$

In agreement with the condition that $|a|<1$ , the angle $\tan^{-1}{a}$ is bound between $-45°$ and $45°$ , i.e. $2\tan^{-1}{a}$ is bound between $-90°$ and $90°$ .

Similarly, ${\frac{2a}{1-a^2}}$ (the main function of the arctangent) is also bound to the same angles.

Therefore, these two angles differ by less than 180°. Thus, equation $A$ is true only when $n+m=0$ .

Hence,

$x=180°n+\tan^{-1}{a},$ $y=-180°n+\tan^{-1}{a}$ , $n\in {Z}$ .

And finally,

$(x-\tan^{-1}{a})^{2015}+(y-\tan^{-1}{a})^{2015}=(180°n+\tan^{-1}{a}-\tan^{-1}{a})^{2015}+(-180°n+\tan^{-1}{a}-\tan^{-1}{a})^{2015}=(180°n)^{2015}-(180°n)^{2015}=\boxed{0}$

Let $\displaystyle b=x+y-2\arctan a$ then $\displaystyle b=\arctan (\frac{2a}{1-a^2}) - 2\arctan a$

Then $\displaystyle \tan b = \tan (\arctan (\frac{2a}{1-a^2}) - 2\arctan a)$

Using formulas $\displaystyle \tan(\alpha -\beta)=\frac{\tan \alpha -\tan \beta}{1+\tan \alpha \tan \beta}$ and $\displaystyle \tan 2\alpha = \frac{2\tan \alpha}{1-\tan^2\alpha}$ , we have:

$\displaystyle \tan b = \frac{\tan (\arctan (\frac{2a}{1-a^2})) - \tan (2\arctan a))}{1-\tan (\arctan (\frac{2a}{1-a^2}))\tan (2\arctan a))}$

$\displaystyle = \frac{\frac{2a}{1-a^2} - \frac{2\tan \arctan a}{1-\tan^2\arctan a}}{1-\tan (\arctan (\frac{2a}{1-a^2}))\tan (2\arctan a))}$

$\displaystyle = \frac{\frac{2a}{1-a^2} - \frac{2a}{1-a^2}}{1-\tan (\arctan (\frac{2a}{1-a^2}))\tan (2\arctan a))}=0$

so, $\displaystyle \tan b=0 <=> b=0$