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It's quite easy and well known that for an angle $0<\alpha<\pi/2$ one has that $$\sin(\alpha)<\alpha,$$ while for cosine one can use the bisection identity to get that $$\frac{1-\cos(\alpha)}{2}=\sin^2\left(\frac{\alpha}{2}\right)<\frac{\alpha^2}{4} \implies 1-\frac{\alpha^2}{2}<\cos(\alpha).$$ This second inequality is pretty nice because it reflects the second order Taylor approximation of cosine, which is quite accurate.

What I would like to have is to obtain the third order inequality for sine, $$\sin(\alpha)>\alpha-\frac{\alpha^3}{6}$$ in a geometric/trigonometric way, i.e. without using calculus. I tried approaching the inequality using the triple angle identity $$\sin(\alpha)=3\sin(\alpha/3)-4\sin^3(\alpha/3)>3\sin(\alpha/3)-4(\alpha/3)^3=3\sin(\alpha/3)-\frac{4\alpha^3}{27}>3\sin(\alpha/3)-\frac{\alpha^3}{6}$$ but obviously I cannot get from here to the correct one.

Any hints? Thanks in advance

b00n heT
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  • The problem I see is that $3x-4x^3$ is an increasing function on this range, so you need a lower bound on $\sin(\alpha/3)$ to proceed, but that is ultimately what you want to get. So you're looking for some other geometric lower bound for $\sin$ on $[0,\pi/6]$ that you can use. Does concavity (which has some geometric interpretation) suffice? In other words, does the bound $\sin(x) \geq 3x/\pi$ suffice? – Ian Nov 02 '19 at 14:49
  • No, that doesn't suffice, you need a bound that's better than that near $x=0$ (since $3\sin(\alpha/3)$ is so close to $\alpha$ for small $\alpha$). – Ian Nov 02 '19 at 14:56
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    See my answer to Evaluate $\lim_{x\to 0}\frac{x-\sin x}{x\sin x}$ Without L'Hopital for a lot of references to this, almost all of which are freely available on the internet. – Dave L. Renfro Nov 02 '19 at 14:59
  • @DaveL.Renfro Thanks a lot for the many references! I think this settles the question then. I'll present a simple answer below soon and then close the question. – b00n heT Nov 02 '19 at 15:06

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