Express $\arccos (\frac 12 \sin x)$ as an algebraic function? I am trying to express $\arccos (\frac 12 \sin x)$ as an algebraic function on the intercal $[0, \frac {\pi}{2})$ . I tried to find this by setting up a triangle with sides $1$ , $x$, and $\sqrt {1-x^2}$ , but I couldn't derive the result from here. The motivation is that if I can find this, I think I can come up with a very good algebraic approximation for $\pi$ . Thanks. Adrian Keister suggested that I put up my strategy of approximating $\pi$ , so here it is: For very small $x$: $\sin x\approx x$ $\sin x \approx \cos \frac {\pi}{2}$ $\cos x \sin x\approx\cos \frac {\pi}{2}$ $\frac 12 \sin 2x \approx\cos \frac {\pi}{2}$ $\pi\approx 2 \arccos (\frac 12 \sin 2x)$ This approximation gets better and better as $x$ approaches $0$ .

I am not enthusiastic about the prospects of finding an "algebraic" expression, since one is asked to solve for the angle $ \ \theta \ $ such that $ \ \cos \theta = \frac{1}{2} \sin x \ $ . (It gives nice answers at $ \ x = 0 \ $ and $ \ x = \frac{\pi}{2} \ , $ but in between... not so much.) Looking at trig identities or the Maclaurin series for arccosine and sine doesn't suggest any tidy results.

The function $ \ \arccos ( \frac{1}{2} \sin x ) \ $ [blue in the graph below] is "pretty well" approximated by $ \ \frac{\pi}{3} + \frac{2}{3 \pi} \cdot (x - \frac{\pi}{2} )^2 \ $ [the red curve] , fitting the vertex and $ \ y-$intercept exactly: the error is scarcely more than 3% anywhere in the first quadrant. But it doesn't look there is going to be a straightforward way to describe the blue curve using elementary functions...

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