So, on Reddit, user Straight-Ad-7750 posted the question,
“Which is bigger without using a calculator: $\frac{\sin(2°)}{2}$ or $\frac{\sin(3°)}{3}$?”
(Note: $2°= 2\cdot \frac{\pi}{180}$ and $3°= 3\cdot \frac{\pi}{180}$.)
There are many ways to prove that one is smaller than the other. These proofs are great because each proof uses one or more useful lemma or commonly known identity. Below we give a proof using only trig identities. Proofs using calculus will be given in part 2.
Spoiler Alerts
The answer is stated in the paragraph below, so if you want to try to figure it out yourself, stop reading now!
Also, it turns out that this is a rather easy question to answer if you are a third or fourth year electrical engineering student because most electrical engineers know about the sinc function defined by $$\mathrm{sinc(x)} = \frac{\sin(x)}{x}.$$ But, let’s just assume that we don’t know about the sinc function and look at the proofs that do not require knowledge about the sinc function.
Trig Proof
We can prove that $\frac{\sin(3°)}{3}$ is smaller just by using trig identities.
Lemma: cos(2°) < cos(1°)
Proof: We will apply the cosine double angle formula, the fact that $\sin^2(1°)\neq0$, and the fact that $1>\cos(1°)>0$ to prove the lemma as follows:
$$\cos(2°) = \cos^2(1°) – \sin^2(1°) < \cos^2(1°) < \cos(1°). □$$
Theorem: $\sin(2°)/2 > \sin(3°)/3$.
Proof: By the fact $1>\cos(1°)>0$, the Lemma, the double angle formula for sine, and the sin of a sum formula,
3/2 = 1 + 1/2
> cos(1°) + 1/2*( cos(2°) / cos(1°))
= cos(1°) + sin(1°)cos(2°) / ( 2 sin(1°) cos(1°))
= cos(1°) + sin(1°)cos(2°) / sin(2°)
= [ sin(2°) cos(1°) + sin(1°)cos(2°)] / sin(2°)
= sin(3°) /sin(2°).
So,
$$3/2 > \sin(3°) /\sin(2°)$$
and hence
$$\sin(2°)/2 > \sin(3°)/3. □$$
Next week I hope to post two calculus based proof. Both of those proofs have cool, useful, simple Lemma!