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# Test of Mathematics Solution Subjective 116 - Angles in a Triangle

This is a Test of Mathematics Solution Subjective 116 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

## Problem

If A, B, C are the angles of a triangle, then show that $\displaystyle { \sin A + \sin B - \cos C \le \frac {3 \sqrt{3}}{2}}$

## Solution:

$\displaystyle { \sin A + \sin B - \cos C }$
$\displaystyle { = \sin A + \sin B - \cos (\pi - (A+B)) }$
$\displaystyle { = \sin A + \sin B + \sin (A+B) }$
$\displaystyle { = 2\sin \frac{(A+B)}{2} \cos \frac{(A-B)}{2} + 2\sin \frac{(A+B)}{2} \cos \frac{(A+B)}{2} }$
$\displaystyle { = 2\sin \frac{(A+B)}{2} \left( \cos \frac{(A-B)}{2} + \cos \frac{(A+B)}{2}\right ) }$
$\displaystyle { = 2\sin \frac{(A+B)}{2} 2\cos \frac{A}{2} \cos \frac{B}{2} }$
$\displaystyle { = 4\sin \frac{(\pi -C)}{2} \cos \frac{A}{2} \cos \frac{B}{2} }$
$\displaystyle { = 4\sin \left(\frac{\pi}{2} - \frac{C}{2} \right) \cos \frac{A}{2} \cos \frac{B}{2} }$
$\displaystyle { = 4\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} }$

We apply Jensen's Inequality and Arithmetic Mean - Geometric Mean inequality here. Since cosine function is concave in the interval $[0, \frac{\pi}{2} ]$, we have
$\displaystyle { \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \frac{\cos \frac{C}{2} + \cos \frac{A}{2} +\cos \frac{B}{2}}{3} \le \cos \left ( \frac{1}{3}\times \frac{A}{2} + \frac{1}{3}\times \frac{B}{2} + \frac{1}{3}\times \frac{C}{2} \right ) }$
This implies $\displaystyle { \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \cos \left ( \frac{A+B+C}{6}\right ) }$
$\displaystyle { \Rightarrow \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \cos \frac{\pi}{6} }$
$\displaystyle { \Rightarrow \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \frac{\sqrt{3}}{2} }$
$\displaystyle { \Rightarrow \cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \le \frac{3\sqrt{3}}{8} }$
$\displaystyle { \Rightarrow 4\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \le \frac{3\sqrt{3}}{2} }$

## Chatuspathi:

• What is this topic: Properties of triangle
• What are some of the associated concept: Jensen Inequality, Arithmetic Mean-Geometric Mean Inequality
• Where can learn these topics:  I.S.I. & C.M.I. Entrance Course of Cheenta, discusses these topics in the ‘Inequality’ module.
• Book Suggestions: 'Secrets in Inequality' by Pam Kim Hung

This is a Test of Mathematics Solution Subjective 116 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

## Problem

If A, B, C are the angles of a triangle, then show that $\displaystyle { \sin A + \sin B - \cos C \le \frac {3 \sqrt{3}}{2}}$

## Solution:

$\displaystyle { \sin A + \sin B - \cos C }$
$\displaystyle { = \sin A + \sin B - \cos (\pi - (A+B)) }$
$\displaystyle { = \sin A + \sin B + \sin (A+B) }$
$\displaystyle { = 2\sin \frac{(A+B)}{2} \cos \frac{(A-B)}{2} + 2\sin \frac{(A+B)}{2} \cos \frac{(A+B)}{2} }$
$\displaystyle { = 2\sin \frac{(A+B)}{2} \left( \cos \frac{(A-B)}{2} + \cos \frac{(A+B)}{2}\right ) }$
$\displaystyle { = 2\sin \frac{(A+B)}{2} 2\cos \frac{A}{2} \cos \frac{B}{2} }$
$\displaystyle { = 4\sin \frac{(\pi -C)}{2} \cos \frac{A}{2} \cos \frac{B}{2} }$
$\displaystyle { = 4\sin \left(\frac{\pi}{2} - \frac{C}{2} \right) \cos \frac{A}{2} \cos \frac{B}{2} }$
$\displaystyle { = 4\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} }$

We apply Jensen's Inequality and Arithmetic Mean - Geometric Mean inequality here. Since cosine function is concave in the interval $[0, \frac{\pi}{2} ]$, we have
$\displaystyle { \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \frac{\cos \frac{C}{2} + \cos \frac{A}{2} +\cos \frac{B}{2}}{3} \le \cos \left ( \frac{1}{3}\times \frac{A}{2} + \frac{1}{3}\times \frac{B}{2} + \frac{1}{3}\times \frac{C}{2} \right ) }$
This implies $\displaystyle { \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \cos \left ( \frac{A+B+C}{6}\right ) }$
$\displaystyle { \Rightarrow \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \cos \frac{\pi}{6} }$
$\displaystyle { \Rightarrow \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \frac{\sqrt{3}}{2} }$
$\displaystyle { \Rightarrow \cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \le \frac{3\sqrt{3}}{8} }$
$\displaystyle { \Rightarrow 4\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \le \frac{3\sqrt{3}}{2} }$

## Chatuspathi:

• What is this topic: Properties of triangle
• What are some of the associated concept: Jensen Inequality, Arithmetic Mean-Geometric Mean Inequality
• Where can learn these topics:  I.S.I. & C.M.I. Entrance Course of Cheenta, discusses these topics in the ‘Inequality’ module.
• Book Suggestions: 'Secrets in Inequality' by Pam Kim Hung