$y''$ in terms of $y$ if $y=\cos^{-1}x$

Question

Find $\frac{d^2y}{dx^2}$ if $y=\cos^{-1}x$ in terms of $y$ only.

The solution is given as $y''=-\cot y.\csc^2y$, but is it the only solution ?

My Attempt

$$ y'=\frac{-1}{\sqrt{1-x^2}} \\y''=\frac{1}{2\sqrt{1-x^2}.(1-x^2)}.-2x=\frac{-x}{(1-x^2)^{{3}/{2}}} $$ Here $y=\cos^{-1}x\implies x=\cos y$ $$ y''=\frac{-\cos y}{(\sin^2y)^{3/2}}=\frac{-\cos y}{|\sin y|^3}=\frac{-\cos y}{|\sin y|}.\frac{1}{\sin^2y}=\color{blue}{\pm\csc^2y.\cot y} $$

Answer 1

I would do it this way: $$x=\cos(y)$$ Differentiate both sides with respect to $x$: $$1=-\sin(y)\frac{\mathrm{d}y}{\mathrm{d}x}$$ $$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{\sin(y)}$$ Now differentiate again: $$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}y}{\mathrm{d}x}\cot(y)\csc(y)$$ $$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\frac{1}{\sin(y)}\cot(y)\csc(y)$$ $$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\frac{\cos(y)}{\sin^3(y)}$$ And @Andreas covered your sign error already, but a little bit of extension:

Osberve that $\text{ran}{(\cos^{-1})}=[0,\pi]$, and the $\sin(x)\geq 0$ iniqualits holds $\forall x \in [0, \pi]$, so you need to pick the positive root. But you can avoid it with using the method I've used.

Note: If you invert the $\cos$ on the interval $[-\pi, 0]$, then the $\sin$ will be negative (or 0) there, so you will need to choose the negative sine. But the usual way of inverting the cosine is on the $[0, \pi]$ interval.

Answer 2

You are introducing an option of signs which does not exist. Start with your previous result here:

$$ y''=\frac{- x}{\sqrt{1-x^2}.(1-x^2)} $$

Observe that the root is only the positive root (!) which is $+ \sin y$, so you get

$$ y''=\frac{-\cos(y)}{\sin^3(y)} = - \csc^2y.\cot y $$

This is the only solution (apart from the usual angle shift properties of the trig functions).

Answer 3

$$x=\cos y,$$ $$1=-y'\sin y,$$ $$0=-y''\sin y-y'^2\cos y,$$

$$\color{green}{y''=-\frac{\cos y}{\sin^3y}}.$$