# Trigonometric Values!

$\large{ \sin(40^\circ) < \sqrt{\dfrac{3}{7}}}$

Prove, without using a calculator, that the above inequality holds true.

Note by Satyajit Mohanty
5 years, 5 months ago

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Suppose 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 $\sin(40^\circ) < \sqrt{\frac37}$.

- 5 years, 5 months ago

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

- 5 years, 5 months ago

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

- 5 years, 5 months ago

Then how did you say that $\sin(x) \leq x - \frac{x^3}6$, or am I missing something?

- 5 years, 5 months ago

Whooops. I forgot to add in one more term. Silly me. Let me fix that.

- 5 years, 5 months ago

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

- 5 years, 5 months ago

Oh. I fixed one small typo. This is what I get for not proofreading my solution. Thanks for correcting me.

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

- 5 years, 5 months ago

I have seen this question just today. I'll think about another solution. But your solution is very good.

- 5 years, 5 months ago

Thanks. I was thinking of a geometric interpretation but I failed at every turn. Anyway, hope to see yours!

- 5 years, 5 months ago

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.

- 5 years, 5 months ago

I tried sum/difference identity and some obviously wrong geometric constructions.

- 5 years, 5 months ago

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

- 5 years, 5 months ago

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 $f \left( \sqrt{\dfrac{3}{7}} \right) > 0$.

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

- 5 years, 5 months ago

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)

- 5 years, 5 months ago

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

- 5 years, 5 months ago

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

- 5 years, 5 months ago

Interesting solution. Can we work from my sin(10) <= 1/7 and use your triple angle formula solution as well?

- 5 years, 5 months ago

Thanks. Maybe we can try substituting for $x$ and arrive at a contradiction.

- 5 years, 5 months ago

aha yes it works!! For $x = \frac\pi{18}$, 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.

By the way, great solution!

- 5 years, 5 months ago

Awesome! I also found a geometric interpretation solution.

- 5 years, 5 months ago

Can't wait to see it!

- 5 years, 5 months ago

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

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

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

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

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

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

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

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

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

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

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

- 5 years, 5 months ago

Typo: "that the above inequality holds true" ;P

- 5 years, 4 months ago

Thanks. Amended. :)

- 5 years, 4 months ago