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Test of Mathematics Solution Subjective 79 -Trigonometric Inequality

This is a Test of Mathematics Solution Subjective 79 (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.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Let ${{\theta}_1}$, ${{\theta}_2}$, ... , ${{\theta}_{10}}$ be any values in the closed interval ${[0,\pi]}$. Show that
${F}$ = ${(1 + {\sin}^2 \theta_1)(1 + {\cos}^2 \theta_1)(1 + {\sin}^2 \theta_2)(1 + {\cos}^2 \theta_2).........(1 + {\sin}^2 \theta_{10})(1 + {\cos}^2 \theta_{10})}$ ${\displaystyle{\le({\frac{9}{4}})^{10}}}$.
What is the maximum value attainable by ${F}$ and at what values of ${{\theta}_1}$, ${{\theta}_2}$, ... , ${{\theta}_{10}}$, is the maximum value attained?

Solution

${F}$ = ${\displaystyle{(1 + {\sin}^2 \theta_1)(1 + {\cos}^2 \theta_1)(1 + {\sin}^2 \theta_2)(1 + {\cos}^2 \theta_2).........(1 + {\sin}^2 \theta_{10})(1 + {\cos}^2 \theta_{10})}}$
Now we will show that for any ${\theta \in}$ ${[0,\pi]}$ ${\displaystyle{(1 + {\sin}^2 \theta)(1 + {cos}^2 \theta) < {\frac{9}{4}}}}$
${\Leftrightarrow}$ ${\displaystyle{2 + {{\sin}^2{\theta}}{{\cos}^2{\theta}}}}$ ${\displaystyle{< {\frac{9}{4}}}}$
${\Leftrightarrow}$ ${\displaystyle{{{\sin}^2{\theta}}{{\cos}^2{\theta}}}}$ ${\displaystyle{< {\frac{1}{4}}}}$
Now ${\displaystyle{{{\sin}^2{\theta}} + {{\cos}^2{\theta}} = 1}}$
${\Rightarrow}$ ${\displaystyle{({{\sin}^2{\theta}} + {{\cos}^2{\theta}})^2 = 1}}$
${\Rightarrow}$ ${\displaystyle{{\sin}^4{\theta} + {\cos}^4{\theta} + 2 {{\sin}^2{\theta}{\cos}^2{\theta}} = 1}}$
${\Rightarrow}$ ${\displaystyle{4 {{\sin}^2{\theta}}{{\cos}^2{\theta}}}}$ ${\displaystyle{< 1}}$ [ as ${{a^2 + b^2} > {2ab}}$ ]
${\Rightarrow}$ ${\displaystyle{{{\sin}^2{\theta}}{{\cos}^2{\theta}}}}$ ${\displaystyle{< {\frac{1}{4}}}}$
So for any ${\theta \in}$ ${[0,\pi]}$ ${\displaystyle{(1 + {\sin}^2 \theta)(1 + {\cos}^2 \theta) {\le} {\frac{9}{4}}}}$
So ${F {\le} ({\frac{9}{4}})^{10}}$ (proved)
Maximum value attained by ${F}$ is ${({\frac{9}{4}})^{10}}$ and will be attained for ${{\theta_i} = {\frac{\pi}{2}} \pm {\frac{\pi}{4}}}$ for ${i = 1, 2, ...., 10}$

This is a Test of Mathematics Solution Subjective 79 (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.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Let ${{\theta}_1}$, ${{\theta}_2}$, ... , ${{\theta}_{10}}$ be any values in the closed interval ${[0,\pi]}$. Show that
${F}$ = ${(1 + {\sin}^2 \theta_1)(1 + {\cos}^2 \theta_1)(1 + {\sin}^2 \theta_2)(1 + {\cos}^2 \theta_2).........(1 + {\sin}^2 \theta_{10})(1 + {\cos}^2 \theta_{10})}$ ${\displaystyle{\le({\frac{9}{4}})^{10}}}$.
What is the maximum value attainable by ${F}$ and at what values of ${{\theta}_1}$, ${{\theta}_2}$, ... , ${{\theta}_{10}}$, is the maximum value attained?

Solution

${F}$ = ${\displaystyle{(1 + {\sin}^2 \theta_1)(1 + {\cos}^2 \theta_1)(1 + {\sin}^2 \theta_2)(1 + {\cos}^2 \theta_2).........(1 + {\sin}^2 \theta_{10})(1 + {\cos}^2 \theta_{10})}}$
Now we will show that for any ${\theta \in}$ ${[0,\pi]}$ ${\displaystyle{(1 + {\sin}^2 \theta)(1 + {cos}^2 \theta) < {\frac{9}{4}}}}$
${\Leftrightarrow}$ ${\displaystyle{2 + {{\sin}^2{\theta}}{{\cos}^2{\theta}}}}$ ${\displaystyle{< {\frac{9}{4}}}}$
${\Leftrightarrow}$ ${\displaystyle{{{\sin}^2{\theta}}{{\cos}^2{\theta}}}}$ ${\displaystyle{< {\frac{1}{4}}}}$
Now ${\displaystyle{{{\sin}^2{\theta}} + {{\cos}^2{\theta}} = 1}}$
${\Rightarrow}$ ${\displaystyle{({{\sin}^2{\theta}} + {{\cos}^2{\theta}})^2 = 1}}$
${\Rightarrow}$ ${\displaystyle{{\sin}^4{\theta} + {\cos}^4{\theta} + 2 {{\sin}^2{\theta}{\cos}^2{\theta}} = 1}}$
${\Rightarrow}$ ${\displaystyle{4 {{\sin}^2{\theta}}{{\cos}^2{\theta}}}}$ ${\displaystyle{< 1}}$ [ as ${{a^2 + b^2} > {2ab}}$ ]
${\Rightarrow}$ ${\displaystyle{{{\sin}^2{\theta}}{{\cos}^2{\theta}}}}$ ${\displaystyle{< {\frac{1}{4}}}}$
So for any ${\theta \in}$ ${[0,\pi]}$ ${\displaystyle{(1 + {\sin}^2 \theta)(1 + {\cos}^2 \theta) {\le} {\frac{9}{4}}}}$
So ${F {\le} ({\frac{9}{4}})^{10}}$ (proved)
Maximum value attained by ${F}$ is ${({\frac{9}{4}})^{10}}$ and will be attained for ${{\theta_i} = {\frac{\pi}{2}} \pm {\frac{\pi}{4}}}$ for ${i = 1, 2, ...., 10}$

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