In part 1, I presented a proof that sin(2°)/2 > sin(3°)/3 using only trigonometry. In part 2, I presented a proof using the Taylor series expansion of sine. I part 3 below, I present three additional proofs: a proof using the average value of a function and the mean value theorem, a proof by Reddit user Dala_The_Pimp which proves $f(x)=\sin(x)/x=\mathrm{sinc}(x)$ is strictly decreasing on $(0,\pi/2)$ using derivatives, and a proof using concavity of $\sin(x)$ on $(0, \pi)$.

It suffices to prove that $f(x) = \sin(x)/x$ is deceasing on $[0, \pi/2]$.

Let $$\mathrm{aveVal}(g(t) , x) = \frac1x \int_0^x g(t) dt.$$

**Lemma** 1: If $g(t)$ is continuous and strictly decreasing on $[0,a]$ and $$f(x)= \mathrm{aveVal}(g(t) , x),$$then $f(x)$ is strictly decreasing on $[0,a]$.

**Proof**: Notice that if $g(t)$ is continuous and $f(x)= \mathrm{aveVal}(g(t) , x)$, then for all $x\in[0,a]$,

$$f'(x) = -\frac1{x^2} \int_0^x g(t) dt + \frac{g(x)}x$$

$$= \frac1{x^2} \left( -\int_0^x g(t) dt + x g(x)\right)$$

$$= \frac1{x^2} ( -x g(c) + x g(x))$$

where $c\in(0,x)$ by the Mean Value Theorem, so

$$f'(x)= \frac1x (g(x) – g(c) ) <0$$

which proves the lemma.

Now,

$$\begin{aligned} \sin(x) – x &= \int_0^x (\cos[t] – 1)\, dt,\mathrm{\ so} \\ \sin(x)/x – 1 &=\mathrm{aveVal}(\cos(t) -1, x).\end{aligned}$$

By Lemma 1, $\cos(t)-1$ is strictly decreasing on $[0, \pi]$ implies

$$\mathrm{aveVal}(\cos(t)-1, 0, x)$$is strictly decreasing on $[0, \pi]$ implies$$\sin(x)/x – 1$$ is strictly decreasing on $[0, \pi]$ implies

$\sin(x)/x$ is strictly decreasing on $[0, \pi]$.

### Reddit post

Below I’m going to copy Reddit user Dala_The_Pimp’s solution Feb 8, 2023. (I am revising it just a little.)

Let $f(x)=\sin(x)/x$. Then

$$f'(x)= \frac{x\cos(x) -\sin(x)}{x^2}.$$

Now let $g(x) = x\cos(x) -\sin(x).$ Clearly $g(0)=0.$ Also

$$g'(x)=-x\sin(x)$$

which is clearly negative in $(0,\pi/2)$ thus $g(x)$ is decreasing hence $g(x)<g(0)$ and consequently $g(x)<0$, thus it is proven that since $g(x)<0, f'(x)<0$ and $f(x)$ is decreasing, hence $f(2^o)>f(3^o)$ and $\sin(2^o)/2^o >\sin(3^o)/3^o$ which implies

$$\sin(2^o)/2 >\sin(3^o)/3.$$

### proof using CONCAVITY

We also can use concavity to prove that $\sin(2^o)/2 >\sin(3^o)/3.$ The second derivative of $\sin(x)$ is $-\sin(x)$, thus the second derivative of $\sin(x)$ is strictly negative on $(0, \pi)$. That implies that $\sin(x)$ is strictly concave on $(0,\pi)$. (See e.g. the proof wiki). By the definition of strict concavity, for all $\lambda\in (0,1)$,

$$\sin( (1-\lambda)\cdot 0 + \lambda 3^o) > (1-\lambda)\sin(0) + \lambda\sin(3^0).$$

Setting $\lambda = 2/3$, we get$$\sin( 2^o) > 2/3 \sin(3^0),\mathrm{\ \ and\ hence}$$ $$\sin( 2^o)/2 > \sin(3^0)/3.$$