Need help in solving

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

tan⁡θ1<sin⁡θ1+sin⁡θ2+sin⁡θ3cos⁡θ1+cos⁡θ2+cos⁡θ3<tan⁡θ3. \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 .

#Geometry

Note by Aman Thegreat
3 years, 8 months ago

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Comments

since,

0<θ1<θ2<θ3<900<sin(θ1)<sin(θ2)<sin(θ3)<11>cos(θ1)>cos(θ2)>cos(θ3)>0Thus, we can saysin(θ1)+sin(θ1)+sin(θ1)<sin(θ1)+sin(θ2)+sin(θ3)<sin(θ3)+sin(θ3)+sin(θ3)cos(θ1)+cos(θ1)+cos(θ1)>cos(θ1)+cos(θ2)+cos(θ3)>cos(θ3)+cos(θ3)+cos(θ3)  ⟹  sin(θ1)+sin(θ1)+sin(θ1)cos(θ1)+cos(θ1)+cos(θ1)<sin(θ1)+sin(θ2)+sin(θ3)cos(θ2)+cos(θ3)+cos(θ3)<sin(θ3)+sin(θ3)+sin(θ3)cos(θ3)+cos(θ3)+cos(θ3)Numerator is an increasing sequence,while denominator is decreasing  ⟹  tan(θ1)<sin(θ1)+sin(θ2)+sin(θ3)cos(θ2)+cos(θ3)+cos(θ3)<tan(θ3)\begin{aligned} 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{#3D99F6} \text{Numerator is an increasing sequence,}\\ &&\small\color{#3D99F6} \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{aligned}

Anirudh Sreekumar - 3 years, 8 months ago

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

Aman thegreat - 3 years, 8 months ago

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sin(θ1)<sin(θ2)(1)sin(θ1)<sin(θ3)(2)(1)+(2) gives,sin(θ1)+sin(θ1)<sin(θ2)+sin(θ3)Adding sin(θ1)to both sidessin(θ1)+sin(θ1)+sin(θ1)<sin(θ1)+sin(θ2)+sin(θ3)\begin{aligned}sin(\theta_1)&<sin(\theta_2)\hspace{5mm}\color{#3D99F6}(1)\\ sin(\theta_1)&<sin(\theta_3)\hspace{5mm}\color{#3D99F6}(2)\\ \color{#3D99F6}(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{aligned}

same logic goes for the other inequalities

Anirudh Sreekumar - 3 years, 8 months ago

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

Aman thegreat - 3 years, 8 months 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 years, 8 months ago

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@Anirudh Sreekumar – Thanks a lot! Brilliant solution .

Aman thegreat - 3 years, 8 months 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.

Mariano PerezdelaCruz - 3 years, 7 months ago
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