0
$\begingroup$

I managed to, in the wild, acquire the function $f(x) = \sin{\frac{\arccos(1-2x)}{2}}$. I checked using Wolfram Alpha for possible simplifications, and found out that it equals (hover to show):

$\sqrt{x}$

I struggled to understand the reasons for this, and discovered this question asking about an equivalent problem (title provides the answer in a different form):

How to prove $2\arccos(x)+\arccos(1-2x^2)=π$ on $x\in[0,1]$ from MVT

But this form of the question seems much more like an exercise than a problem-solving step-by-step method. How can one discover this relationship without prior knowledge of it (either from empirical graphed data or from an exercise)?

$\endgroup$
1
  • 1
    $\begingroup$ Use round brackets rather than curly ones around a mathematical function's argument, as in my edit. Curly brackets, or braces, are for arguments of MathJaX functions such as \operatorname. $\endgroup$
    – J.G.
    May 15, 2020 at 20:47

1 Answer 1

3
$\begingroup$

Let $\theta:=\arccos(1-2x)$ so $x=\frac{1-\cos\theta}{2}=\sin^2\frac{\theta}{2}$ and $\sin\frac{\theta}{2}=\pm\sqrt{x}$. But an arccosine $\in[0,\,\pi]$, so $\sin\frac{\theta}{2}\ge0$ and $\sin\frac{\theta}{2}=\sqrt{x}$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .