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
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?
${F}$ = $ {\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
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?
${F}$ = $ {\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}$
