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

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

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

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

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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 \(\sin(40^\circ) < \sqrt{\frac37} \). – Pi Han Goh · 1 year, 3 months ago

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– Ishan Singh · 1 year, 3 months ago

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

– Pi Han Goh · 1 year, 3 months ago

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

– Ishan Singh · 1 year, 3 months ago

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

– Pi Han Goh · 1 year, 3 months ago

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

– Ishan Singh · 1 year, 3 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 \(\sin x \geq x - \frac {x^3}{6}\)Log in to reply

By the way, do you have a solution of your own? I do not like my solution. – Pi Han Goh · 1 year, 3 months ago

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– Ishan Singh · 1 year, 3 months ago

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

– Pi Han Goh · 1 year, 3 months ago

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

– Ishan Singh · 1 year, 3 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.Log in to reply

– Pi Han Goh · 1 year, 3 months ago

I tried sum/difference identity and some obviously wrong geometric constructions.Log in to reply

– Lakshya Sinha · 1 year, 3 months ago

Nice solution but I did not understand a single thing, explanation please and a reference link too.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. – Ishan Singh · 1 year, 3 months ago

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

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

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And it's obvious that sin(2pi/9) > 1/2 because sin(2pi/9) > sin(pi/6) – Pi Han Goh · 1 year, 3 months ago

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– Ishan Singh · 1 year, 3 months ago

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

– Ishan Singh · 1 year, 3 months ago

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

– Pi Han Goh · 1 year, 3 months ago

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

– Ishan Singh · 1 year, 3 months ago

Thanks. Maybe we can try substituting for \(x\) and arrive at a contradiction.Log in to reply

By the way, great solution! – Pi Han Goh · 1 year, 3 months ago

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– Ishan Singh · 1 year, 3 months ago

Awesome! I also found a geometric interpretation solution.Log in to reply

– Pi Han Goh · 1 year, 3 months ago

Can't wait to see it!Log in to reply

Typo:"that the aboveinequalityholds true" ;P – Karthik Venkata · 1 year, 3 months agoLog in to reply

– Satyajit Mohanty · 1 year, 2 months ago

Thanks. Amended. :)Log in to reply