which is larger: $\sin(4^\circ)$ or $2\sin(2^\circ)$? I had two expression that simplified to the ones in the title.
Obviously, I can't use a calculator.
We didn't learn the double angle (or half angle) formulas so Ill have to find a different way, maybe with the unit circle.
 A: Here is a simple way to see this. The 
chord function is defined as
$$\text{crd}(x) := 2\sin(x/2).$$
In order to show that $$\sin(2x)<2\sin(x)$$ 
for all $\,x\,$ it is
equivalent to showing that $$\text{crd}(2x)<2\,\text{crd}(x)$$ for all $\,x\,$ but this
is just the triangle inequality applied to an isoceles triangle inscribed in a 
circle. The base of the triangle is the chord of
twice the angle of the other two equal sides which
are chords the the same angle.
A: The sine wave is concave down between $0$ and $180^\circ$, so the line that passes through the origin and $(2^\circ,\sin 2^\circ)$ will pass above $(4^\circ,\sin4^\circ)$.  Thus, $\sin 4^\circ<2\sin 2^\circ$.
A: hint: What's the length of the red line? 

A: Use Taylor expansion:
$$\sin x=x-\frac{x^3}{3!}+O(x^5)\\
\sin 4^\circ <\frac{3.14}{45}-\frac16\left(\frac{3.14}{45}\right)^3+\frac1{120}\left(\frac{3.14}{45}\right)^5<2\left(\frac{3.14}{90}-\frac16\left(\frac{3.14}{90}\right)^3\right)<2\sin 2^\circ. $$
A: Attempt:
$f(x):= \sin (2x) -2 \sin x ;$ $f(0)=0$;
$f'(x)=2\cos (2x) -2\cos x=$
$2(\cos (2x)-\cos x)<0$.
Hence $f(2°) <0$.
Note $\cos x$ is strictly decreasing
in $(0,π/2)$ ($(\cos x)'=-\sin x <0)$.
