Arctan in 2015

Just use the formula - $arctan x + arctan y = arctan(\frac{x + y}{1 - xy})$

And of course, $arctan x - arctan y = arctan(\frac{x - y}{1 + xy})$

Io, here you have the answer, then.

Rewrite as

$\arctan 2 + \arctan 0 + \arctan 1 + \arctan 5 = \arctan 2015 + \arctan x.$

First of all, $\arctan 0 = 0,$ so we are left with

$\arctan 2 + \arctan 1 + \arctan 5 = \arctan 2015 + \arctan x.$

To solve this problem, we use complex numbers. The complex number with argument (angle with respect to the real axis) $\arctan 2$ is $1 + 2i,$ since the tangent of the argument is $\frac{2}{1}.$ Similarly, we find that $1 + i,$ $1 + 5i,$ and $1 + 2015i$ have arguments $\arctan 1,$ $\arctan 5,$ and $\arctan 2015,$ respectively.

Let the complex number with argument $\arctan x$ be $a + bi,$ so that $x = \frac{b}{a}.$ The complex number with argument $\arctan 2 + \arctan 1 + \arctan 5$ is $(1 + 2i)(1 + i)(1 + 5i),$ while the complex number with argument $\arctan 2015 + \arctan x$ is $(1 + 2015i)(a + bi).$ Setting these two equal and solving for $a + bi$ yields

$\begin{aligned} (1 + 2i)(1 + i)(1 + 5i) &= (1 + 2015i)(a + bi) \\ -16 - 2i &= (1 + 2015i)(a + bi) \\ a + bi &= -\frac{16 + 2i}{1 + 2015i} \\ a + bi &= -\frac{4046 - 32238i}{1^2 + 2015^2}. \end{aligned}$

Finally,

$\begin{aligned} x &= \frac{b}{a} \\ &= \frac{(1^2 + 2015^2)b}{(1^2 + 2015^2)a} \\ &= \frac{32238}{-4046} \\ &= -\frac{16119}{2023} \end{aligned}$

and $p + q = 16119 + 2023 = \boxed{18142}.$

NOTE: Even though the value of $x$ we found in this solution does not satisfy

$\arctan 2 + \arctan 0 + \arctan 1 + \arctan 5 = \arctan 2015 + \arctan x,$

it does satisfy the equation posed in the problem, since there is no range restriction on $x$ .

Let $\alpha = \arctan 2$ , $\beta = \arctan 1$ , $\gamma = \arctan 0$ , $\delta = \arctan 5$ and $\epsilon = \arctan 2015$ .

Then, we have:

$\tan(\arctan 2 + \arctan 1 + \arctan 0 + \arctan 5 - \arctan 2015) \\ = \tan (\alpha + \beta + \color{#3D99F6}{\gamma} + \delta - \epsilon) \quad \quad \quad \small \color{#3D99F6}{\text{We note that } \gamma = \arctan 0 = 0} \\ = \tan (\alpha + \beta + \delta - \epsilon) \\ = \dfrac{\tan (\alpha+\beta) + \tan (\delta -\epsilon)}{1-\tan (\alpha+\beta) \tan (\delta -\epsilon)} \quad \quad \small \color{#3D99F6}{\tan (\alpha+\beta) = \dfrac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta} = \dfrac{2+1}{1-2} = - 3 \quad \tan (\delta -\epsilon) = - \dfrac{1005}{5038}} \\ = \dfrac{3-\frac{1005}{5038}}{1-\frac{3(1005)}{5038}} \\ = - \dfrac{16119}{2023}$

$\Rightarrow p + q = 16119 + 2023 = \boxed{18142}$

None of the solutions are correct. On the left side you have an angle greater than π, and on the right side you have an angle smaller than π/2 with another angle subtracted from it, which is clearly impossible. (Read Steven's solution first and continue) Here is the reason: graph the point a+bi=-(4046-32238i)/(1+2015^2), it lies in Quadrant II, which means that arg(a+bi)>π/2. Hence its argument cannot be the output of an arctangent function, and arg(a+bi)≠arctan(b/a). If you put in arctan(b/a), you get the angle directly opposite arg(a+bi) since they have the same tangent.