# Need help in solving

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
8 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}

- 8 months, 1 week ago

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

- 8 months, 1 week ago

\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

- 8 months, 1 week ago

After that, how did you divide the inequalities ?

- 8 months, 1 week ago

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

- 8 months, 1 week ago

Thanks a lot! Brilliant solution .

- 8 months, 1 week ago

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.

- 7 months ago