Math god needed for a trig problem I concted

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hw

Delete this, please. I'm stoopid.
 
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That's not particularly special: cos of something is a number between 0 and 1, so you're taking sin of something small (less than 1 degree). So things obviously converge very quickly.
 
Do a search for small angle approximation
 
The above posters hit the nail on the head. In fact, sin(cos(tan(sin(cos(tan(sin(cos(tan(n), is not equal for any integer n. The difference is just so small that your calculator can not display the difference. Thus, the small angle approximation.

Try entering sin(cos(tan(sin(cos(tan(1) and sin(cos(tan(sin(cos(tan(1E10). If your calculator is capable of displaying enough digits, you will find that the approximate value of the first is .017452406437037, while the approximate value of the second is .017452406437039.

If your calculator can't display that many digits, use sin(cos(tan(n) instead.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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