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

Question

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.

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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?

Answer 1

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 $x\le \frac{\pi}2$.

Answer 2

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%

Answer 3

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.

Answer 4

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 5

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.

Answer 6

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.

Answer 7

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.