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Assume that we only know $\tan (0)=0$ and also given the relation $\tan'(x)=1+\tan^2(x)$ about $\tan (x)$ and we do not know other $\tan (x)$ relations of trigonometry.

How can I get the additon formula $$ \tan (x+h)=\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)}$$ via using the differential equation ($\tan'(x)=1+\tan^2(x)$) and analytic methods ? (without using geometry)

Could you please provide me with an easy way to get addition formula of $\tan (x+h)$ ?

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My attempt:

$$\tan'(x)=1+\tan^2(x)$$

$$\int \frac{d\tan(x)}{1+\tan^2(x)}=\int dx$$

$$\tan(x)- \frac{\tan^3(x)}{3}+ \frac{\tan^5(x)}{5}+….=x$$

$$\tan(h)- \frac{\tan^3(h)}{3}+ \frac{\tan^5(h)}{5}+….=h$$

$$+$$

$$\tan(x)+\tan(h)- (\frac{\tan^3(x)}{3}+\frac{\tan^3(h)}{3})+ (\frac{\tan^5(x)}{5}+\frac{\tan^5(h)}{5})+….=x+h$$

$$\tan(x+h)- \frac{\tan^3(x+h)}{3}+ \frac{\tan^5(x+h)}{5}+….=x+h$$

$$\tan(x+h)- \frac{\tan^3(x+h)}{3}+ \frac{\tan^5(x+h)}{5}+….=\tan(x)+\tan(h)- (\frac{\tan^3(x)}{3}+\frac{\tan^3(h)}{3})+ (\frac{\tan^5(x)}{5}+\frac{\tan^5(h)}{5})+….=x+h$$

Let’s define that $\tan(x+h)=\tan(x)+\tan(h)+P(x,h)$

$$(\tan(x)+\tan(h)+P(x,h))- \frac{(\tan(x)+\tan(h)+P(x,h))^3}{3}+ \frac{(\tan(x)+\tan(h)+P(x,h))^5}{5}+….=\tan(x)+\tan(h)- (\frac{\tan^3(x)}{3}+\frac{\tan^3(h)}{3})+ (\frac{\tan^5(x)}{5}+\frac{\tan^5(h)}{5})+….$$

$$P(x,h)- \frac{(\tan(x)+\tan(h))^3+ 3 (\tan(x)+\tan(h))^2 P(x,h)+3 (\tan(x)+\tan(h)) P(x,h)^2+P(x,h)^3}{3}+ \frac{(\tan(x)+\tan(h)+P(x,h))^5}{5}+….=- (\frac{\tan^3(x)}{3}+\frac{\tan^3(h)}{3})+ (\frac{\tan^5(x)}{5}+\frac{\tan^5(h)}{5})+….$$

Let’s define that

$$P(x,h)= \tan(x)\tan(h)(\tan(x)+\tan(h))+G(x,h) $$

thus we get

$\tan(x+h)=\tan(x)+\tan(h)+\tan(x)\tan(h)(\tan(x)+\tan(h))+G(x,h)$

I know we can find the solution in my method but so many calculations are needed in that method. It is very very long way.

And Finally I need to get that $$ \tan (x+h)=\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)}=(\tan(x)+\tan(h))(1-\tan(x)\tan(h))^{-1}=(\tan(x)+\tan(h))(1+\tan(x)\tan(h)+\tan^2(x)\tan^2(h)+……)$$

Note:I try to get addition formulas from a given differential equations and initial conditions.

I wish to find a method to get a closed form addition formulas for the problem as shown below.

$U(0)=0$ and also given the relation $U'(x)=1+U^n(x)$ where $n>2$

Thanks a lot for answers

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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}

$$

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