# 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.

hint: What's the length of the red line? • John.Nice answer. Dec 29, 2019 at 16:22
• Thank you for your kind words, Peter, but I thought that everybody had great answers. :-) Dec 31, 2019 at 14:58

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.

• Hipparchus would be proud. :-) Dec 29, 2019 at 16:23

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

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

• I must say, that's a great solution haha too bad we're still in fundamental trigonometry Dec 29, 2019 at 20:29

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 $$(0,π/2)$$ ($$(\cos x)'=-\sin x <0)$$.