Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

\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?

share|cite|improve this question
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
up vote 5 down vote accepted

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






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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.