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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 $\tan (x+h)$ ?

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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Let $\phi(x,h)$ be $\frac{\tan(x)+\tan(h)}{1 - \tan(x)\tan(h)}$, you can verify $\phi$ satisfies the same ODE $\frac{d}{dx} y(x) = 1 + y(x)^2$ like $\tan(x)$. Since $\phi(x,h) = \tan(x+h)$ at $x = 0$, you can use the fundamental theorem of ODE (in particular, the uniqueness part of solution to an ODE ) to conclude $\phi(x,h) = \tan(x+h)$. –  achille hui Feb 1 '13 at 15:16
    
@achillehui : We assume that we do not know $$ \tan (x+h)=\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)}$$ . Of course ,I know it will satify same differential equation. Could you estimate addition formula for $U'(x)=1+U^3(x)$? –  Mathlover Feb 1 '13 at 15:25
    
You can always start with a study on what happens to the ODE under various transformations. If you try that with Mobius transformation, you will find that some parameters will leave your ODE invariant and lead you to above sum rule. No idea what need to be done to $U'(x) = 1 + U^3(x)$. –  achille hui Feb 1 '13 at 15:55
    
Do we know $\tan(x) = \frac{\sin(x)}{\cos(x)}$, $\sin^2(x) + \cos^2(x) = 1$ and $\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$? –  Pragabhava Feb 19 '13 at 18:58
    
@Pragabhava : No. I know very well how to get addition formula of tan via trigonometric relations. We assume that we only know differential equation. My aim is to find an analytic way to solve general problem as I wrote in my question. I think the method that you wanted me to do 2000clicks.com/mathhelp/GeometryTrigEquivTanSum.aspx –  Mathlover Feb 20 '13 at 6:52
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1 Answer

up vote 7 down vote accepted
+500

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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Thank you very much for your both solution. Have you tried to solve the generalized problem ($U'=1+U^n$) via using your methods? Is it possible to find closed forms addition formulas for general problem? Thanks a lot again for your help. –  Mathlover Feb 22 '13 at 7:37
    
@Mathlover I added other examples. The method works with Euler equations $$\frac{dx}{\sqrt{X}} + \frac{dy}{\sqrt{Y}} = 0,$$ where \begin{align}{}X &= a_0x^4 + a_1 x^3 + a_2 x^2 + a_3 x + a_4\\ Y &= a_0y^4 + a_1 y^3 + a_2 y^2 + a_3 y + a_4\end{align} It strongly relies on the construction of the second solution. For the problem $y' = 1 + y^n$, this doesn't seem to be straight forward, but maybe some modification can be attempted. –  Pragabhava Feb 22 '13 at 19:00
2  
Lovely answer. The colored equations were a nice touch. (+1) –  Evan Teitelman Feb 22 '13 at 19:43
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