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

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestsince,

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

Log in to reply

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

Log in to reply

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

same logic goes for the other inequalities

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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.

Log in to reply