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...

## 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