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# 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
1 year, 8 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 $$\sin(40^\circ) < \sqrt{\frac37}$$. · 1 year, 8 months ago

Just a query, how will you show that the remaining terms in the Maclaurin series are negative? The terms after $$\ldots$$ · 1 year, 8 months ago

I did'nt say that it's negative, but sum of every two consecutive terms are positive. · 1 year, 8 months ago

Then how did you say that $$\sin(x) \leq x - \frac{x^3}6$$, or am I missing something? · 1 year, 8 months ago

Whooops. I forgot to add in one more term. Silly me. Let me fix that. · 1 year, 8 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 $$\sin x \geq x - \frac {x^3}{6}$$ · 1 year, 8 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. · 1 year, 8 months ago

I have seen this question just today. I'll think about another solution. But your solution is very good. · 1 year, 8 months ago

Thanks. I was thinking of a geometric interpretation but I failed at every turn. Anyway, hope to see yours! · 1 year, 8 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. · 1 year, 8 months ago

I tried sum/difference identity and some obviously wrong geometric constructions. · 1 year, 8 months ago

Nice solution but I did not understand a single thing, explanation please and a reference link too. · 1 year, 8 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

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

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

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

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

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

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

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

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

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

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

Contradiction. · 1 year, 7 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 $$f \left( \sqrt{\dfrac{3}{7}} \right) > 0$$.

$$\therefore \dfrac{1}{2} < \sin \left(\dfrac{2\pi}{9}\right) < \sqrt{\dfrac{3}{7}}$$. · 1 year, 8 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) · 1 year, 8 months ago

$$f \left( \sqrt{\dfrac{3}{7}} \right) = \sqrt{3} \left( 1 - \left(\dfrac{18}{7\sqrt{7}}\right) \right)$$ and since $$(7\sqrt{7})^2 > 18^2$$, $$f \left( \sqrt{\dfrac{3}{7}} \right) > 0$$ · 1 year, 8 months ago

Yeah, that's obvious. I just wrote it to illustrate the location of the root. · 1 year, 8 months ago

Interesting solution. Can we work from my sin(10) <= 1/7 and use your triple angle formula solution as well? · 1 year, 8 months ago

Thanks. Maybe we can try substituting for $$x$$ and arrive at a contradiction. · 1 year, 8 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! · 1 year, 8 months ago

Awesome! I also found a geometric interpretation solution. · 1 year, 7 months ago

Can't wait to see it! · 1 year, 7 months ago

Typo: "that the above inequality holds true" ;P · 1 year, 7 months ago