This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .

Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.

Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

@Pi Han Goh
–
I get that sum of every two consecutive terms must be positive. I think I got confused due to the typo. It should be $\sin x \geq x - \frac {x^3}{6}$

@Pi Han Goh
–
What all did you try in geometric interpretation? I wrote $\sin \left( \dfrac{2\pi}{9} \right) = \displaystyle \int_{0}^{\frac{2\pi}{9}} \cos x \ \mathrm{d} x$ and tried Cauchy-Schwarz inequality for integrals with different functions, but it didn't work out well enough to prove the desired inequality.

Consider the polynomial $f(x) = 8x^3 - 6x + \sqrt{3}$. The roots of $f(x)$ are $\sin \left(\dfrac{\pi}{9}\right), \sin \left(\dfrac{2\pi}{9}\right)$ and $-\sin \left(\dfrac{4\pi}{9}\right)$, the largest root being $\sin \left(\dfrac{2\pi}{9}\right)$. Note that, $f(0) > 0$, $f \left(\dfrac{1}{2} \right) < 0$ and $f \left( \sqrt{\dfrac{3}{7}} \right) > 0$.

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestSuppose otherwise, that is suppose that $\sin(40^\circ) \geq \sqrt{\frac37}$, then squaring both sides gives $\sin^2(40^\circ) \geq \frac 37 \Rightarrow 1 - 2\sin^2(40^\circ) \leq -\frac67 + 1 = \frac17$, equivalently by double angle formula, $\cos(80^\circ) = \sin(10^\circ) = \sin\left(\frac\pi{18}\right) \leq \frac17$. By Maclaurin Series, $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{120} \ldots$, so $\sin(x) \leq x - \frac{x^3}6 + \frac{x^5}{120}$. Thus we have

$\frac{\pi}{18} - \frac16 \left( \frac{\pi}{18} \right)^3 + \frac1{120} \left( \frac\pi{18}\right)^5 \leq \frac17$

$\pi (120\cdot18^4 - 18^2 \cdot\pi^2 \cdot 20 + \pi^4) \leq \frac{120\cdot18^5}7$

$3 \times 18^2\cdot20 (6\cdot18^2-\pi) < \frac{120\cdot18^5}7$

$120\cdot 3 \cdot18^4 < \frac{120\cdot18^5}7$

$3 < \frac{18}7$

which is clearly absurd, thus $\sin(40^\circ) < \sqrt{\frac37}$.

Log in to reply

Just a query, how will you show that the remaining terms in the Maclaurin series are negative? The terms after $\ldots$

Log in to reply

I did'nt say that it's negative, but sum of every two consecutive terms are positive.

Log in to reply

$\sin(x) \leq x - \frac{x^3}6$, or am I missing something?

Then how did you say thatLog in to reply

Log in to reply

$\sin x \geq x - \frac {x^3}{6}$

I get that sum of every two consecutive terms must be positive. I think I got confused due to the typo. It should beLog in to reply

By the way, do you have a solution of your own? I do not like my solution.

Log in to reply

Log in to reply

Log in to reply

$\sin \left( \dfrac{2\pi}{9} \right) = \displaystyle \int_{0}^{\frac{2\pi}{9}} \cos x \ \mathrm{d} x$ and tried Cauchy-Schwarz inequality for integrals with different functions, but it didn't work out well enough to prove the desired inequality.

What all did you try in geometric interpretation? I wroteLog in to reply

Log in to reply

Nice solution but I did not understand a single thing, explanation please and a reference link too.

Log in to reply

Consider the polynomial $f(x) = 8x^3 - 6x + \sqrt{3}$. The roots of $f(x)$ are $\sin \left(\dfrac{\pi}{9}\right), \sin \left(\dfrac{2\pi}{9}\right)$ and $-\sin \left(\dfrac{4\pi}{9}\right)$, the largest root being $\sin \left(\dfrac{2\pi}{9}\right)$. Note that, $f(0) > 0$, $f \left(\dfrac{1}{2} \right) < 0$ and $f \left( \sqrt{\dfrac{3}{7}} \right) > 0$.

$\therefore \dfrac{1}{2} < \sin \left(\dfrac{2\pi}{9}\right) < \sqrt{\dfrac{3}{7}}$.

Log in to reply

How do you know that f(sqrt(3/7)) > 0?

And it's obvious that sin(2pi/9) > 1/2 because sin(2pi/9) > sin(pi/6)

Log in to reply

$f \left( \sqrt{\dfrac{3}{7}} \right) = \sqrt{3} \left( 1 - \left(\dfrac{18}{7\sqrt{7}}\right) \right)$ and since $(7\sqrt{7})^2 > 18^2$, $f \left( \sqrt{\dfrac{3}{7}} \right) > 0$

Log in to reply

Yeah, that's obvious. I just wrote it to illustrate the location of the root.

Log in to reply

Log in to reply

$x$ and arrive at a contradiction.

Thanks. Maybe we can try substituting forLog in to reply

$x = \frac\pi{18}$, we got $-4\sin^3(x) + 3\sin(x) = \frac12\Rightarrow -4\sin^3(x) \geq \frac12 -\frac37 = \frac1{14} \Rightarrow \sin^3(x) \leq - \frac1{14\cdot4} < 0$ which is clearly absurd.

aha yes it works!! ForBy the way, great solution!

Log in to reply

Log in to reply

Log in to reply

Consider a triangle with angles $A$, $B$ and $C$ such that $A=\dfrac{4\pi}{9}$, $B = \dfrac{2\pi}{9}$ and $C = \dfrac{\pi}{3}$ . We have,

$\sin \left(\dfrac{A}{2}\right) \cdot \sin \left(\dfrac{B}{2}\right) \cdot \sin \left(\dfrac{C}{2}\right) < \dfrac{1}{8}$

$\implies \sin \left(\dfrac{\pi}{9}\right) \sin\left(\dfrac{2\pi}{9}\right) < \dfrac{1}{4}$

Assume, on the contrary, that

$\sin \left(\dfrac{2\pi}{9}\right) \geq \sqrt{\dfrac{3}{7}}$

$\implies \sin \left(\dfrac{\pi}{18}\right) \leq \dfrac{1}{7}$

$\implies \cos \left(\dfrac{4\pi}{9}\right) \leq \dfrac{1}{7}$

$\implies \tan\left(\dfrac{4\pi}{9}\right) \geq 4\sqrt{3}$

$\implies \tan\left(\dfrac{2\pi}{9}\right) \geq \dfrac{\sqrt{3}}{2}$

$\implies \tan\left(\dfrac{2\pi}{9}\right) \geq \sin\left(\dfrac{\pi}{3}\right)$

$\implies 2\sin \left(\dfrac{2\pi}{9}\right) \geq 2\sin\left(\dfrac{\pi}{3}\right) \cos\left(\dfrac{2\pi}{9}\right)$

$\implies 2\sin \left(\dfrac{2\pi}{9}\right) \geq \sin \left(\dfrac{4\pi}{9}\right) + \sin \left(\dfrac{\pi}{9}\right)$

$\implies 2\sin \left(\dfrac{2\pi}{9}\right) \geq 2\sin \left(\dfrac{2\pi}{9}\right) \cos \left(\dfrac{2\pi}{9}\right) + \sin \left(\dfrac{\pi} {9}\right)$

$\implies \sin \left(\dfrac{\pi}{9}\right) \sin\left(\dfrac{2\pi}{9}\right) \geq \dfrac{1}{4}$

Contradiction.

Log in to reply

Typo:"that the aboveinequalityholds true" ;PLog in to reply

Thanks. Amended. :)

Log in to reply