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If \( 0^\circ < \theta_1 < \theta_2 < \theta_3 < 90^\circ \), prove that

\[ \tan \theta_1 < \dfrac{ \sin \theta_1 + \sin \theta_2 + \sin \theta_3 }{ \cos \theta_1 + \cos \theta_2 + \cos \theta_3 } < \tan \theta_3 .\]

Note by Aman Thegreat
3 months, 2 weeks ago

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since,

\(\begin{align} 0&<\theta_1<\theta_2<\theta_3<90\\ 0<sin(\theta_1)&<sin(\theta_2)<sin(\theta_3)<1\\ 1>cos(\theta_1)&>cos(\theta_2)>cos(\theta_3)>0\\ \text{Thus, we can say}\\\\ sin(\theta_1)+sin(\theta_1)+sin(\theta_1)&<sin(\theta_1)+sin(\theta_2)+sin(\theta_3)<sin(\theta_3)+sin(\theta_3)+sin(\theta_3)\\\\ cos(\theta_1)+cos(\theta_1)+cos(\theta_1)&>cos(\theta_1)+cos(\theta_2)+cos(\theta_3)>cos(\theta_3)+cos(\theta_3)+cos(\theta_3)\\\\ \implies\dfrac{sin(\theta_1)+sin(\theta_1)+sin(\theta_1)}{cos(\theta_1)+cos(\theta_1)+cos(\theta_1)}&<\dfrac{sin(\theta_1)+sin(\theta_2)+sin(\theta_3)}{cos(\theta_2)+cos(\theta_3)+cos(\theta_3)}<\dfrac{sin(\theta_3)+sin(\theta_3)+sin(\theta_3)}{cos(\theta_3)+cos(\theta_3)+cos(\theta_3)}\hspace{5mm}&\small\color{blue} \text{Numerator is an increasing sequence,}\\ &&\small\color{blue} \text{while denominator is decreasing}\\ \implies tan(\theta_1)&<\dfrac{sin(\theta_1)+sin(\theta_2)+sin(\theta_3)}{cos(\theta_2)+cos(\theta_3)+cos(\theta_3)}<tan(\theta_3)\end{align}\)

Anirudh Sreekumar - 3 months, 2 weeks ago

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Hey, can you explain the first and second step after "thus, we can say" ..?

Aman Thegreat - 3 months, 2 weeks ago

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\(\begin{align}sin(\theta_1)&<sin(\theta_2)\hspace{5mm}\color{blue}(1)\\ sin(\theta_1)&<sin(\theta_3)\hspace{5mm}\color{blue}(2)\\ \color{blue}(1)+(2) \text{ gives,}\\ sin(\theta_1)+sin(\theta_1)&<sin(\theta_2)+sin(\theta_3)\\\\ \text{Adding } sin(\theta_1) \text{to both sides}\\ sin(\theta_1)+sin(\theta_1)+sin(\theta_1)&<sin(\theta_1)+sin(\theta_2)+sin(\theta_3)\end{align}\)

same logic goes for the other inequalities

Anirudh Sreekumar - 3 months, 2 weeks ago

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@Anirudh Sreekumar After that, how did you divide the inequalities ?

Aman Thegreat - 3 months, 2 weeks ago

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@Aman Thegreat For the fractions, each of the numerators is greater than the last,each of the denominators is less than the previous one. so the fractions are increasing

Anirudh Sreekumar - 3 months, 2 weeks ago

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@Anirudh Sreekumar Thanks a lot! Brilliant solution .

Aman Thegreat - 3 months, 2 weeks ago

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I am not good at formatting so I just am giving you a hint. Make Theta 1 = Theta 2 - Delta Theta ; Make Theta 3 = Theta 2 + Delta Theta. Bearing in mind that Theta 2 is arbitrary and Delta Theta is an infinitesimal quantity, therefore Cos ( Delta Theta)=1 and Sin( Delta Theta)=0. By substitution in the fraction and expanding the sum a sines and cosines you will get that the value of the fraction is tangent of Theta 2 which is in agreement with the proof. Of course geometrically if you compare the segment which represents tangents you will see is function which increase value up to 90 degree which became infinite.

Mariano PerezdelaCruz - 2 months, 1 week ago

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