which is larger: $\sin(4^\circ)$ or $2\sin(2^\circ)$?

Question

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.

Answer 1

hint: What's the length of the red line?

Answer 2

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.

Answer 3

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, $\sin 4^\circ<2\sin 2^\circ$.

Answer 4

Use Taylor expansion: $$\sin x=x-\frac{x^3}{3!}+O(x^5)\\ \sin 4^\circ <\frac{3.14}{45}-\frac16\left(\frac{3.14}{45}\right)^3+\frac1{120}\left(\frac{3.14}{45}\right)^5<2\left(\frac{3.14}{90}-\frac16\left(\frac{3.14}{90}\right)^3\right)<2\sin 2^\circ. $$

Answer 5

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 $(0,π/2)$ ($(\cos x)'=-\sin x <0)$.