Angular order and elementary inverse-trigonometric bounds

The cosine assertion is immediate from [F1]. On $[0,\pi \mathrm{/}2]$ sine increases from $0$ to $1$, and on $[\pi \mathrm{/}2,\pi ]$ it decreases from $1$ to $0$, so it is nonnegative on the whole interval. By the inverse definition and these endpoint values, $\arcsin 0=0$, $\arcsin 1=\pi \mathrm{/}2$, and $\arcsin$ maps $[0,1]$ into $[0,\pi \mathrm{/}2]$.

Put $f(t)=\arcsin t$. Its derivative $f^{\mathrm{\prime }}(t)=1\mathrm{/}\sqrt{1-t^2}$ is strictly increasing on $[0,1)$: for $0\le s<t<1$, the positive square roots satisfy $\sqrt{1-s^2}>\sqrt{1-t^2}$. For $0<x<1$, apply the mean value theorem on $[0,x]$ and $[x,1]$. The resulting interior points $a<x<b$ give $f(x)\mathrm{/}x=f^{\mathrm{\prime }}(a)<f^{\mathrm{\prime }}(b)=(f(1)-f(x))\mathrm{/}(1-x)$, and hence $f(x)<xf(1)=\pi x\mathrm{/}2$. Equality holds at both endpoints by step 1.1.

If $0<x\le 1\mathrm{/}4$, the mean value theorem on $[0,x]$ gives $f(x)=x\mathrm{/}\sqrt{1-c^2}$ for some $0<c<x$. Thus $f(x)\le 4x\mathrm{/}\sqrt{15}<21x\mathrm{/}20=1.05x$: the strict numerical comparison follows by squaring the positive sides, since $1600<1653$. At $x=0$ equality holds.

Let $g(x)=\log (1+x)-2x\mathrm{/}(2+x)$ for $x\ge 0$. It is continuous on every closed interval $[0,x]$, differentiable inside, and $g(0)=0$. The derivative rules give $g^{\mathrm{\prime }}(t)=1\mathrm{/}(1+t)-4\mathrm{/}(2+t{)}^2=t^2\mathrm{/}((1+t)(2+t{)}^2)$ for $t>0$. This derivative is positive, so the mean value theorem gives $g(x)>0$ for $x>0$; at $x=0$ it is zero. These results establish all the claims.