# The tangent half-angle substitution in differential equations

\begin{align} y & = \tan\frac\theta2 \\[8pt] \frac{1-y^2}{1+y^2} & = \cos\theta \\[8pt] \frac{2y}{1+y^2} & = \sin\theta \\[8pt] \frac{2\,dy}{1+y^2} & = d\theta \end{align} This, the tangent half-angle substitution, is famously used in solving equations of the form $$\frac{df}{d\theta} = f(\cos\theta,\sin\theta)$$ where $f$ is a rational function. I.e. it is used for finding antiderivatives of rational functions of sine and cosine.

Is the same substitution used in solving any other, more elaborate, differential equations?

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Did you have something in mind? –  Ron Gordon Jan 6 '13 at 23:06
@rlgordonma : No. I've extrapolated from this substitution in a different and apparently novel direction; this is another direction that I haven't really thought about. –  Michael Hardy Jan 6 '13 at 23:39
One generalisation of this is into algebraic geometry, where this kind of parametrisation of a curve (in this case a circle from a point on the circumference, based at root - as I understand it - on angle at centre is twice angle at circumference) becomes rational map/birational isomorphism at a level of abstraction rather less than currently encountered in such courses. Which goes to say, I don't know the answer ... –  Mark Bennet Jan 6 '13 at 23:48

In fact the Weierstrass substitution can have usage in ODEs, for example is to transform a linear ODE of trigonometric function coefficients to a linear ODE of polynomial function coefficients whose letting $u=\sin kx$ or $u=\cos kx$ cannot work.

For example $(a_1\sin x+b_1\cos x+c_1)\dfrac{d^2y}{dx^2}+(a_2\sin x+b_2\cos x+c_2)\dfrac{dy}{dx}+(a_3\sin x+b_3\cos x+c_3)y=0~:$

Let $u=\tan\dfrac{x}{2}$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=\dfrac{1}{2}\left(\sec^2\dfrac{x}{2}\right)\dfrac{dy}{du}=\dfrac{1}{2}\left(\tan^2\dfrac{x}{2}+1\right)\dfrac{dy}{du}=\dfrac{u^2+1}{2}\dfrac{dy}{du}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{u^2+1}{2}\dfrac{dy}{du}\right)=\dfrac{d}{du}\left(\dfrac{u^2+1}{2}\dfrac{dy}{du}\right)\dfrac{du}{dx}=\left(\dfrac{u^2+1}{2}\dfrac{d^2y}{du^2}+u\dfrac{dy}{du}\right)\dfrac{u^2+1}{2}=\dfrac{(u^2+1)^2}{4}\dfrac{d^2y}{du^2}+\dfrac{u(u^2+1)}{2}\dfrac{dy}{du}$

$\therefore\left(\dfrac{2a_1u}{u^2+1}-\dfrac{b_1(u^2-1)}{u^2+1}+c_1\right)\left(\dfrac{(u^2+1)^2}{4}\dfrac{d^2y}{du^2}+\dfrac{u(u^2+1)}{2}\dfrac{dy}{du}\right)+\left(\dfrac{2a_2u}{u^2+1}-\dfrac{b_2(u^2-1)}{u^2+1}+c_2\right)\dfrac{u^2+1}{2}\dfrac{dy}{du}+\left(\dfrac{2a_3u}{u^2+1}-\dfrac{b_3(u^2-1)}{u^2+1}+c_3\right)y=0$

$\dfrac{2a_1u-b_1(u^2-1)+c_1(u^2+1)}{u^2+1}\left(\dfrac{(u^2+1)^2}{4}\dfrac{d^2y}{du^2}+\dfrac{u(u^2+1)}{2}\dfrac{dy}{du}\right)+\dfrac{2a_2u-b_2(u^2-1)+c_2(u^2+1)}{u^2+1}\dfrac{u^2+1}{2}\dfrac{dy}{du}+\dfrac{2a_3u-b_3(u^2-1)+c_3(u^2+1)}{u^2+1}y=0$

$(2a_1u-b_1(u^2-1)+c_1(u^2+1))\left((u^2+1)^2\dfrac{d^2y}{du^2}+2u(u^2+1)\dfrac{dy}{du}\right)+2(2a_2u-b_2(u^2-1)+c_2(u^2+1))(u^2+1)\dfrac{dy}{du}+4(2a_3u-b_3(u^2-1)+c_3(u^2+1))y=0$

$(2a_1u-b_1(u^2-1)+c_1(u^2+1))(u^2+1)^2\dfrac{d^2y}{du^2}+2(2a_1u^2-b_1u(u^2-1)+c_1u(u^2+1)+2a_2u-b_2(u^2-1)+c_2(u^2+1))(u^2+1)\dfrac{dy}{du}+4(2a_3u-b_3(u^2-1)+c_3(u^2+1))y=0$

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+1 for this nice and complete answer. –  Babak S. Jan 10 '13 at 13:52