Express $x=\theta-\sin(\theta)$ using differential notation.

Question

So, I have a curve, $C_1$, defined by the parametric equation $$x=\theta-\sin(\theta),~~~y=1-\cos(\theta)$$(essentially a cycloid with $r=1$), and for proofing purposes, I'm trying to express $x$ using differential notation. I know that typically, we can express functions in differential notation as $$y=f(x)$$ $$\implies dy=f'(x)~dx$$but I'm trying to go about a more concrete way to reach this conclusion.

We start with $$x=\theta-\sin(\theta)$$ $$\implies x+dx=(\theta+d\theta)-\sin(\theta+d\theta)$$ $$dx=(\theta+d\theta)-\sin(\theta+d\theta)-(\theta-\sin(\theta))$$ $$dx=d\theta+\sin(\theta)-\sin(\theta+d\theta)$$ then, using the difference of sines identity, which states that $$\sin(x)-\sin(y)=2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)$$we have that $$dx=d\theta+2\cos\left(\frac{\theta+\theta+d\theta}{2}\right)\sin\left(\frac{\theta-\theta-d\theta}{2}\right)$$ $$dx=d\theta-2\cos\left(\theta+\frac{d\theta}{2}\right)\sin\left(\frac{d\theta}{2}\right)$$and from here I've expanded out the cosine using the angle addition formulae, but, and I kid you not, after further manipulation, that got me straight back to where I started. So I'll save you the hassle of typing out each and every step until then, but what I hope to have happen is that I'll end up with $$dx=d\theta-d\theta\cos(\theta)$$ $$\implies dx=(1-\cos(\theta))d\theta$$and since $x'=1-\cos(\theta)$, $$\implies dx=x'd\theta$$which would mean we have successfully expressed $x$ using differential notation. Obviously, seeing as I'm here, though, I haven't quite reached this step. Can anyone guide me down the garden path? Thank you.

Any help is appreciated.

Answer 1

You are correct until this step:

$$dx=d\theta-2\cos\left(\theta+\frac{d\theta}{2}\right)\sin\left(\frac{d\theta}{2}\right)$$

After this you need to note that:

1. $\cos(x+dx)\approx\cos(x)$
2. $\sin(dx)\approx dx$

both of which are standard concepts taught when manipulating trigonometric functions in limits. Using these, we get:

$$\begin{align}dx&=d\theta-2\cos\left(\theta\right)\left(\frac{d\theta}{2}\right)\\&=d\theta(1-\cos\theta)\end{align}$$

which is as expected.