Convergent series with trigonometric functions like $a_{0}=m, a_{2n-1}=\sin(a_{2n-2}), a_{2n}=\cos(a_{2n-1})$

Question

I have found by computation that $$a_{0}=m, a_{2n-1}=f(a_{2n-2}), a_{2n}=g(a_{2n-1})$$ converges when $m$ - real and $f,g$ - some trigonometric function. $$(f,g)\to(|\lim\limits_{n\to\infty}a_{2n-1}|, |\lim\limits_{n\to\infty}a_{2n}|)$$ So we have those pairs: $$(\sin,\cos)\to(\approx0,694819, \approx0,768169)$$ $$(\sin,\tan)\to(\approx0,999906, \approx1,557086)$$ $$(\sin,\sec)\to(\approx0,976783, \approx1,786699)$$ $$(\sin,\csc)\to(\approx0,944039, \approx1,234669)$$ $$(\cos,\sec)\to(\approx0,446048, \approx1,108452)$$ $$(\tan,\csc)\to(\approx1,557701, \approx1,000086)$$ $$(\cot,\sec)\to(\approx0,477049, \approx1,125678)$$ And two special cases when they are the same: $$(\cos)\to(\approx0,739085)$$ $$(\csc)\to(\approx1,114157)$$ Why those series converges (or have limit) to some results? What does they mean (relation with some constants)?

Answer 1

Look at this calculation on Wolfram Alpha. Do you see how it explains your results? Do you see why this must be so if the sequences converge? This doesn't prove convergence, of course, but it tells you what the limits must be if the sequences converge.