# Prove that sin(x) ≥ x/2, but without calculus!

## Important Note:

After when this puzzle was posted, many people pointed out errors and improvements that could be made. I also noticed many flaws, so the post once had gone through drastic changes. However, I learned and understood that such behavior is discouraged. I'll leave the puzzle as it was before the flaws were and edited, to make sure the answers and comments make sense for future readers.

The title explains it all. Can you prove that $$\sin(x) \ge x/2$$ holds where $$0 \le x \le \pi/2$$, but without using any calculus?

The only prerequisite that you can regard as true without proving is that on a plane, if a shape Y can be completely covered by a shape X, then X has a larger area than Y.

Notes:

• The theorems you incorporate should not have been derived from calculus, so it would be safe to use most formulas of trigonometry.
• The title of this puzzle was inspired by this challenge.
• Hints and my own solution will be posted after some time passes.
• How far can you increase this lower bound, without the aid of calculus?
• Can you not just compare the graphs of $\sin{x}$ and $x/2$? That's just algebra, no calculus. Sep 8, 2021 at 16:23
• "The only prerequisite that you can regard as true without proving [...]" is absolutely unworkable. There is a vast body of mathematics that you are already assuming by even writing "sin(x)", all of which has to be accepted for the puzzle even to be meaningful. Sep 8, 2021 at 17:01
• My definition of the sine function is that it the solution of a certain differential equation. What is yours? Sep 9, 2021 at 9:46
• That's a major edit, as far as my too-tired-to-math brain can tell. We generally discourage edits which change requirements/specifications, especially if any answers are invalidated by the edit. Here's one meta answer of many Sep 11, 2021 at 5:09
• @bobble I now understand how those edits can confuse readers and might negatively act to answerers. I rollbacked to the revision before. Sep 11, 2021 at 6:57

I think this is as simple as I could get it. In the picture, the shaded sector has area $$\frac{x}2$$ because the full unit circle has area $$\pi$$, making the sector area $$\frac{x}{2\pi}$$ times as much.
The triangle $$ABC$$ has height $$\sin x$$, and base $$|AB| = 2\cos x$$. Its area is therefore $$\sin x\cos x = \frac{\sin {2x}}2$$.
The triangle covers the sector, so $$\sin {2x} = 2A_{ABC} \ge 2A_{sector} = x$$.

This works as long as $$\angle ACB \ge \frac{\pi}2$$, which is when $$x\le \frac{\pi}4$$. By substituting $$x/2$$ for $$x$$ we get the result: $$\sin {x} \ge \frac{x}2$$ for $$x\le \frac{\pi}2$$.

• Could you explain why the triangle covers the sector? Isn't that only true for x <= pi / 4? Sep 8, 2021 at 8:54
• @loopywalt Yes, I was already working on fixing that. Sep 8, 2021 at 9:09
• This is really nice. +1 Sep 8, 2021 at 9:56
• How can you prove (without calculus) that sin(x)*cos(x)=sin(2*x)/2 ? Sep 8, 2021 at 18:27
• @RazvanSocol You can reflect point C across AB to get the point C'. Then calculate the area of triangle AC'C in two ways - base $2\sin x$ with height $\cos x$ or base $1$ with height $\sin 2x$. This is essentially how loopywalt's answer avoids the use of the double angle formula. Sep 8, 2021 at 21:37

Consider the following diagram:

In this image, it is easy to see that the area of sector BAC is equal to $$\alpha/2$$. Furthermore, if the length AC is 1, then length CF is $$\sin \alpha$$ - therefore, since the length AB is also 1, the area of rectangle ABED is also $$\sin \alpha$$.
It is obvious that rectangle ABED fully covers sector BAC, and therefore $$\sin\alpha\geq \alpha/2$$.

To improve on the lower bound using the same model, one option is to

construct another sector in triangle ACD at C - it's easy to confirm that the radius will be $$\cos\alpha$$ and the angle will be $$\alpha$$, so the added area will be $$\frac{\alpha}2\cos^2\alpha$$. Adding this to the existing area, we have $$\sin\alpha\geq \frac\alpha2 + \frac\alpha2\cos^2\alpha=\frac\alpha2(2-\sin^2\alpha)$$
Rearranging, we get $$\sin^2\alpha +\frac2\alpha\sin\alpha - 2 \geq 0$$ Using the quadratic formula to process from here, along with knowledge of quadratics and the fact that $$\sin\alpha<1$$, we obtain $$\sin\alpha\geq \frac{-2/\alpha \pm \sqrt{4/\alpha^2+8}}2$$ Interestingly, this produces a better lower bound, even for $$\alpha=\pi/2$$, despite the geometry of the two models being identical (as the added sector will have area zero).

Here is a plot of the sin function (blue), the original lower bound of $$\alpha/2$$ (green), and the updated lower bound (red): The new lower bound has a maximum relative error of 20%

• Amazing! It's so obvious, so why didn't I see it?! Sep 9, 2021 at 6:26
• This should be the accepted answer. Haha Sep 9, 2021 at 9:22
• "It is obvious that rectangle ABED fully covers sector BAC, and therefore sinα≥α/2" I am wary of any geometric proof that says something is obvious, just because it looks true on one example picture. In particular, this would be false if α wasn't in range [0, pi/2]. So it's really not that obvious at all and in fact requires to be quite careful about the value of α.
– Stef
Sep 9, 2021 at 15:05
• @Stef - it's obvious because $\alpha$ is known to be between 0 and $\pi/2$. This single piece of information, plus the image provided, allows one to quickly see that constructing the line perpendicular to AB through C forms two lines that bound the sector vertically, and defining points D and E by drawing lines perpendicular to AB through A and B and intersecting with the first line provides lines AD and BE that bound the sector horizontally - as shown in the image (the construction process, that is). To put it another way, you're constructing ABED to cover the sector. Sep 10, 2021 at 0:34
• "It is obvious because it depends on an assumption which I didn't mention in my whole post" is a complete self-contradiction. If you had simply said "Since we know that 0 <= α <= pi/2, it follows that rectangle ABED fully covers sector BAC", I wouldn't have commented. But saying that a statement is obvious because it is true on a picture, when in fact that statement is false for most values of α, is a very misleading way to write math proofs. In fact, since you don't explicitly use the assumption 0 <= α <= pi/2 in your whole proof, you effectively proved that sin(x) >= x/2 for all values of x.
– Stef
Sep 10, 2021 at 8:52

Here is a purely trigonometric one:

$$\sin x = 2 \cos (\frac x 2) \sin (\frac x 2) = 2 \cos^2 (\frac x 2) \tan (\frac x 2) \ge 2 \cos^2 (\frac \pi 4) \frac x 2 = \frac x 2$$

This uses the angle doubling formula for the sine and the facts that in the first quadrant (argument between $$0$$ and $$\frac \pi 2$$) the cosine is an absolutely decreasing function and the tangent is at least as large as its argument.

• Can you include an elementary proof that $\tan(x) \geq x$ for $x\in [0,\pi/2)$? On the surface, it seems to be a very similar question to the one being asked here. Sep 9, 2021 at 8:06
• @AlexJones Obvious from definition of tangent in terms of tangent to unit circle (compare area of circular sector to area of containing right triangle). Sep 9, 2021 at 9:48 In the figure the thick black lines are the quantities to compare. sin x is the height of the teal triangle (which therefore has area 1/2 sin x) and x/2 is the length of the circular arc delimiting the purple circular segment (which therefore has area x/4). We need to show that the teal area is more than the purple area. The purple shape is fully contained in the triangle given by orange outline. As this triangle has the same base as the teal triangle it will suffice to compare heights. The blue line connecting tips of the triangles is the mirror image of the height of the orange triangle wrt the long side of the orange triangle. The claim now follows from this side's angle over the base being no larger than pi / 4.

Increasing the bound:

From this proof it is obvious that the bound can be increased from x/2 to tan (x/2).

Answer (sorry for the crude drawing):

Let the angle $$AOB$$ on a unit circle be equal to $$x$$ (in radians), so the $$BH$$ will be $$\sin x$$ by definition. So, the area of the $$AOB$$ triangle will be exactly $$\frac12 \sin x$$ (since the base $$OA=1$$, and the height $$BH=\sin x$$). On the other side, the area of the $$AOB$$ circular sector is $$\frac x2$$. So, that means that we have to prove that the $$AOB$$ triangle has greater area than the AB segment (between the arc and the chord). (It follows from the fact that if $$\sin x \geqslant \frac x2$$, then $$\frac12 \sin x \geqslant \frac x2 - \frac12 \sin x$$ and vice versa.)
The $$AOB$$ triangle is isosceles (because $$OA=OB=1$$), and its height (not $$BH$$, but one originating from $$O$$) is $$\cos\frac x2$$ (since we can divide this triangle into 2 right-angled ones with one of the acute angles being $$\frac x2$$). So, the segment height is $$1-\cos\frac x2$$. That means that we have to prove that $$\cos \frac x2 > 2(1-\cos\frac x2)$$ (since the area of a segment is less than one of a rectangle with the same base and height).
But, when $$0\leqslant x\leqslant \frac\pi2$$, we have $$\cos \frac x2 \geqslant \frac{\sqrt2}2>0.7$$. Since $$0.7>2(1-0.7)=0.6$$ (and it's obvious that the inequality will hold for larger values of $$\cos \frac x2$$, which will make the left-hand side greater, and the right-hand side less), that's all. QED. • The second part of your proof seems overly complicated. I would just draw tangents at A and B to get a triangle that covers the segment, and then show that triangle is smaller than OAB by comparing the base angles. Sep 8, 2021 at 7:39

A look at the sine function (the knowledge of which, I suppose, is a prerequisite for asking the question) shows that the line segment of $$f(x) = \frac{x}{2}$$ is below the sinus wave in that interval because the starting point for both functions is (0,0) but the end point of the line segment is below the sine curve, and the sine curve in this interval is concave, i.e., bulges upward.

The diagram reveals that the slope of the line can be increased exactly until the upper endpoint of the line comes to rest on the sine curve. That would be the point $$(\frac{\pi}{2},1)$$ which results in a slope of $$\frac{2}{\pi}$$, a bit more than 0.6 instead of 0.5.

Admittedly, this is more a strong plausibility than a proof: In particular, there is probably no rigorous definition of "concave" without derivation, I don't show that $$sin(x)$$ actually is concave in that interval, and I don't even prove that from these properties it follows that the curve is above $$x/2$$.

But still, the graph is striking evidence.

My original solution turned out to be same as the top one, so I had to change it:

Even $$sin(π/2)$$ exceeds half of the covered portion of the circle. The narrower the angle becomes, the shorter the distance between the intersection of the perpendicular and right radius and the end of the radius, so the premise is even more true for smaller angles.