To get addition formula of $\tan (x)$ via analytic methods

Solution 1:

Here is my approach:

Let $$ \arctan(x) := \int_0^x \frac{dt}{1+t^2} $$

and $\tan(x)$ is defined as the inverse of $\arctan(x)$, so that $\tan(0) = 0$.

Consider the differential equation $$ \frac{d x}{1+x^2} + \frac{dy}{1+y^2} = 0 \tag{DE}, $$ one solution is $$ \arctan x + \arctan y = c, $$ but the equation has also the solution $$ \frac{x + y}{1 - x y} = C. $$ Since the differential equation has but one distinct solution, the two solutions must be related to one another in a definite way. This relation is expressed by the equation $$ C = f(c) $$ Now, let $$ x = \tan u, \quad y = \tan v, $$ then \begin{align} u + v &= c \\ \\ \frac{\tan u + \tan v}{1 - \tan u \tan v} &= f(c) = f(u +v) \end{align}

Let $v = 0$, then $$ \tan u = f(u) $$ and therefore $$ \color{blue}{\frac{\tan u + \tan v}{1 - \tan u \tan v} = \tan(u + v).} $$

Construction of the second solution

Let $x = \tan u$ and $y = \tan v$. By definition $$ \frac{d x}{d u} = 1 + x^2 \quad \Longrightarrow \quad \frac{d^2 x}{d u^2} = 2x(1+x^2). $$

Similarly $$ \frac{d y}{d u} = -\frac{d y}{d v} = -(1+y^2), \mbox{ and } \frac{d^2 y}{d u^2} = \frac{d^2 y}{d v^2} = 2y(1+y^2) $$

from which follows that $$ x \frac{d^2 y}{d u^2} - y \frac{d^2 x}{d u^2} = 2xy(y^2 - x^2) $$ and $$ x^2\left(\frac{d y}{d u}\right)^2 - y^2 \left(\frac{d x}{d u}\right)^2 = (x^2 - y^2)(1 - x^2 y^2) $$

Hence $$ \frac{x \frac{d^2 y}{d u^2} - y \frac{d^2 x}{d u^2}}{x \frac{d y}{d u} - y \frac{d x}{d u}} = - \left(x \frac{d y}{d u} + y \frac{d x}{d u}\right) \frac{2 x y}{1-x^2y^2} $$

This equation is immediately integrable; the solution is

$$ \log\left(x \frac{d y}{d u} - y \frac{d x}{d u}\right) = \mbox{const.} + \log(1 - x^2 y^2) $$

or $$ x \frac{d y}{d v} + y \frac{d x}{d u} = C(1- x^2 y^2). $$

Using this information, we can see that $$ \Phi(x,y) = \frac{x(1+y^2) + y(1+x^2)}{1 - x^2 y^2} = \frac{x + y}{1 - x y} = C $$ is also a solution of (DE).

Other Examples

• $\color{green}{\sin(x)}$

Let $y' = \sqrt{1-y^2}$. If we consider the (DE) $$ \frac{d x}{\sqrt{1-x^2}} + \frac{d y}{\sqrt{1-y^2}} = 0 $$ and define $$ \arcsin(x) = \int_0^x \frac{d t}{\sqrt{1-t^2}}dt $$ where $\sin(x)$ is defined as the inverse of $\arcsin(x)$, so that $\sin(0) = 0$, and $\cos(x)$ is defined as $\sqrt{1-\sin^2 (x)}$, with the condition that $\cos(0) = 1$.

One solution for the (DE) is $$ \arcsin x + \arcsin y = c. $$

Using the same method as with $\tan(x)$, we can build a second solution:

Let $x = \sin u$, $y = \sin v$, by definition $$ \frac{d x}{d u} = \sqrt{1-x^2}, \qquad \frac{d^2 x}{d u^2} = -x $$ Similarly $$ \frac{d y}{d u} = -\frac{d y}{d u} = -\sqrt{1-y^2}, \qquad \frac{d^2 y}{d u^2} = -y $$ from which follows $$ x \frac{d^2 y}{d u^2} - y \frac{d^2 x}{d u^2} = 0 $$

Hence $$ \frac{x \frac{d^2 y}{d u^2} - y \frac{d^2 x}{d u^2}}{x \frac{d y}{d u} - y \frac{d x}{d u}} = 0 $$ Integrating $$ x \frac{d y}{d v} + y \frac{d x}{d u} = C $$ and another solution of the (DE) is $$ x\sqrt{1-y^2} + y \sqrt{1-x^2} = C $$

Recapitulating:

The (DE) has two solutions $$ \arcsin x + \arcsin y = c, $$ $$ x\sqrt{1-y^2} + y \sqrt{1-x^2} = C $$ As with $\tan(x)$, the (DE) has but one solution, and the two solutions must be related to one another in a definite way $f(c) =C$.

Let $x = \sin u$ and $y = \sin v$, then $$ u + v = c $$ $$ \sin u \cos v + \sin v \cos u = f(c) = f(u+v) $$ Setting $v = 0$ implies $f(u) = \sin u$ so $$ \color{blue}{\sin u \cos v + \sin v \cos u = \sin(u +v)} $$

• $\color{green}{\mbox{sn}(x)}$

Let $y' = (1-y^2)^{\frac{1}{2}}(1-k^2y^2)^{\frac{1}{2}}$. The (DE) $$ \frac{dx}{(1-x^2)^{\frac{1}{2}}(1-k^2x^2)^{\frac{1}{2}}} +\frac{dy}{(1-y^2)^{\frac{1}{2}}(1-k^2y^2)^{\frac{1}{2}}} = 0 $$ has the solutions (using the same technique) $$ \mbox{argsn } x + \mbox{argsn } y = c $$ $$ x \frac{d y}{d v} + y \frac{d x}{d u} = C(1-k^2 x^2 y^2) $$ Let $x = \mbox{sn }u$, $y = \mbox{sn }v$, then $$ \color{blue}{\mbox{sn}(u+v) = \frac{\mbox{sn }u \,\mbox{sn}'v + \mbox{sn }v \,\mbox{sn}'u}{1-k^2 \mbox{sn}^2u \,\mbox{sn}^2v}} $$

• $\color{green}{\wp(x)}$

Let $y' = \sqrt{4x^3-g_2x-g_3}$. In this case, the (DE) is of the form $$ \frac{dx}{\sqrt{4x^3-g_2x-g_3}} + \frac{dy}{\sqrt{4y^3-g_2y-g_3}} $$ and we can derive the addition formula $$ \color{blue}{\wp(u + v) = - \wp(u) - \wp(v) + \frac{1}{4} \left\{\frac{\wp'(u)-\wp'(v)}{\wp(u)-\wp(v)} \right\}^2} $$