Question Number 213084 by MATHEMATICSAM last updated on 30/Oct/24
$$\mathrm{If}\:\mathrm{0}\:<\:{x}\:<\:\frac{\pi}{\mathrm{2}}\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{sin}\left(\mathrm{cos}{x}\right)\:<\:\mathrm{cos}{x}\:<\:\mathrm{cos}\left(\mathrm{sin}{x}\right). \\ $$$$\mathrm{not}\:\mathrm{using}\:\mathrm{graph}. \\ $$
Answered by issac last updated on 30/Oct/24
![x=acos(z) , z∈[0,1] sin(cos(cos^(−1) (z)))<cos(cos^(−1) (z))<cos(sin(cos^(−1) (z))) sin(z)<z<cos((√(1−z^2 ))) sin(z)<z<cos((√(1−z^2 ))) , z∈(0,1] sin(z)≤z ((sin(z))/z)≤1 → ((d )/dz)sin(z)≤1 cos(z)≤1 , for all z∈[0,1] sin(z)≤z≤cos((√(1−z^2 ))) z≤cos((√(1−z^2 ))) e^z =1+z+(1/(2!))z^2 +(1/(3!))z^3 ..... (Tailer series) Couchy−Schwartz Inequality (a_1 z_1 +a_2 z_2 +.......a_n z_n )^2 ≤(a_1 ^2 +a_2 ^2 +...a_n ^2 )(z_1 ^2 +z_2 ^2 +...z_n ^2 ) (1+z+(1/(2!))z^2 +(1/(3!))z^3 .....)^2 ≤(1^2 +1^2 +((1/(2!)))^2 +((1/(3!)))^2 +...)(1^2 +z^2 +z^4 +....) e^(2z) ≤((I_0 (2))/(1−z^2 )) , I_0 (2)≈2.2796.... I_ν (x) is Modified 1st Bessel function e^z ≤((I_0 (2))/(1−(1/4)z^2 )) → e^z ≤((4∙I_0 (2))/(4−z^2 )) , z∈[0,1] e^(iz) +e^(−iz) ≤((2I_0 (2))/((1/4)z^2 +1)) → cos(z)≤((4∙I_0 (2))/(z^2 +4)) cos((√(1−z^2 )))≤((4I_0 (2))/(5−z^2 )) z≤((4I_0 (2))/(5−z^2 )) → 5z−z^3 ≤4I_0 (2) 0≤z^3 −5z+4I_0 (2) satisfy for all z∈[0,1] Q.E.D](https://www.tinkutara.com/question/Q213089.png)
$${x}=\mathrm{acos}\left({z}\right)\:,\:{z}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{sin}\left(\mathrm{cos}\left(\mathrm{cos}^{−\mathrm{1}} \left({z}\right)\right)\right)<\mathrm{cos}\left(\mathrm{cos}^{−\mathrm{1}} \left({z}\right)\right)<\mathrm{cos}\left(\mathrm{sin}\left(\mathrm{cos}^{−\mathrm{1}} \left({z}\right)\right)\right) \\ $$$$\mathrm{sin}\left({z}\right)<{z}<\mathrm{cos}\left(\sqrt{\mathrm{1}−{z}^{\mathrm{2}} }\right) \\ $$$$\mathrm{sin}\left({z}\right)<{z}<\mathrm{cos}\left(\sqrt{\mathrm{1}−{z}^{\mathrm{2}} }\right)\:,\:{z}\in\left(\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{sin}\left({z}\right)\leq{z}\: \\ $$$$\frac{\mathrm{sin}\left({z}\right)}{{z}}\leq\mathrm{1}\:\rightarrow\:\:\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\mathrm{sin}\left({z}\right)\leq\mathrm{1}\:\:\mathrm{cos}\left({z}\right)\leq\mathrm{1}\:,\:\mathrm{for}\:\mathrm{all}\:{z}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{sin}\left({z}\right)\leq{z}\leq\mathrm{cos}\left(\sqrt{\mathrm{1}−{z}^{\mathrm{2}} }\right) \\ $$$${z}\leq\mathrm{cos}\left(\sqrt{\mathrm{1}−{z}^{\mathrm{2}} }\right) \\ $$$${e}^{{z}} =\mathrm{1}+{z}+\frac{\mathrm{1}}{\mathrm{2}!}{z}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}!}{z}^{\mathrm{3}} …..\:\left(\mathrm{Tailer}\:\mathrm{series}\right) \\ $$$$\mathrm{Couchy}−\mathrm{Schwartz}\:\mathrm{Inequality} \\ $$$$\left({a}_{\mathrm{1}} {z}_{\mathrm{1}} +{a}_{\mathrm{2}} {z}_{\mathrm{2}} +…….{a}_{{n}} {z}_{{n}} \right)^{\mathrm{2}} \leq\left({a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} +…{a}_{{n}} ^{\mathrm{2}} \right)\left({z}_{\mathrm{1}} ^{\mathrm{2}} +{z}_{\mathrm{2}} ^{\mathrm{2}} +…{z}_{{n}} ^{\mathrm{2}} \right) \\ $$$$\left(\mathrm{1}+{z}+\frac{\mathrm{1}}{\mathrm{2}!}{z}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}!}{z}^{\mathrm{3}} …..\right)^{\mathrm{2}} \leq\left(\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}!}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{3}!}\right)^{\mathrm{2}} +…\right)\left(\mathrm{1}^{\mathrm{2}} +{z}^{\mathrm{2}} +{z}^{\mathrm{4}} +….\right) \\ $$$${e}^{\mathrm{2}{z}} \leq\frac{{I}_{\mathrm{0}} \left(\mathrm{2}\right)}{\mathrm{1}−{z}^{\mathrm{2}} }\:,\:{I}_{\mathrm{0}} \left(\mathrm{2}\right)\approx\mathrm{2}.\mathrm{2796}…. \\ $$$${I}_{\nu} \left({x}\right)\:\mathrm{is}\:\mathrm{Modified}\:\mathrm{1st}\:\mathrm{Bessel}\:\mathrm{function} \\ $$$${e}^{{z}} \leq\frac{{I}_{\mathrm{0}} \left(\mathrm{2}\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}{z}^{\mathrm{2}} }\:\:\rightarrow\:\:{e}^{{z}} \leq\frac{\mathrm{4}\centerdot{I}_{\mathrm{0}} \left(\mathrm{2}\right)}{\mathrm{4}−{z}^{\mathrm{2}} }\:,\:{z}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$${e}^{\boldsymbol{{i}}{z}} +{e}^{−\boldsymbol{{i}}{z}} \leq\frac{\mathrm{2}{I}_{\mathrm{0}} \left(\mathrm{2}\right)}{\frac{\mathrm{1}}{\mathrm{4}}{z}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:\:\mathrm{cos}\left({z}\right)\leq\frac{\mathrm{4}\centerdot{I}_{\mathrm{0}} \left(\mathrm{2}\right)}{{z}^{\mathrm{2}} +\mathrm{4}} \\ $$$$\mathrm{cos}\left(\sqrt{\mathrm{1}−{z}^{\mathrm{2}} }\right)\leq\frac{\mathrm{4}{I}_{\mathrm{0}} \left(\mathrm{2}\right)}{\mathrm{5}−{z}^{\mathrm{2}} } \\ $$$${z}\leq\frac{\mathrm{4}{I}_{\mathrm{0}} \left(\mathrm{2}\right)}{\mathrm{5}−{z}^{\mathrm{2}} }\:\rightarrow\:\mathrm{5}{z}−{z}^{\mathrm{3}} \leq\mathrm{4}{I}_{\mathrm{0}} \left(\mathrm{2}\right) \\ $$$$\mathrm{0}\leq{z}^{\mathrm{3}} −\mathrm{5}{z}+\mathrm{4}{I}_{\mathrm{0}} \left(\mathrm{2}\right)\:\:\mathrm{satisfy}\:\mathrm{for}\:\mathrm{all}\:{z}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathbb{Q}.\mathbb{E}.\mathbb{D} \\ $$
Commented by issac last updated on 30/Oct/24
$$\mathrm{I}\:\mathrm{use}\:\mathrm{Couchy}−\mathrm{schwartz}\:\mathrm{inequality} \\ $$$$\mathrm{and}\:\mathrm{Manipulated}\:\mathrm{inequality} \\ $$$$\mathrm{if}\:\mathrm{A},\mathrm{B}\:\mathrm{is}\:\mathrm{positive}\:\mathrm{and}\:{k}\:\mathrm{is}\:\:\mathrm{positive} \\ $$$$\mathrm{if}\:\mathrm{A}\leq\mathrm{B}\:,\:\:\mathrm{A}+{k}\leq\mathrm{B}+{k}\:\mathrm{is}\:\mathrm{true} \\ $$$$\left.{S}\mathrm{o},\:\mathrm{I}\:\mathrm{use}\:\mathrm{that}\:\mathrm{fact}\::\right)\: \\ $$
Answered by mr W last updated on 30/Oct/24
$${for}\:\mathrm{0}<{x}<\frac{\pi}{\mathrm{2}}: \\ $$$$\mathrm{sin}\:{x}<{x}\:\Rightarrow\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)>\mathrm{cos}\:{x}\:\checkmark \\ $$$$\mathrm{0}<\mathrm{cos}\:{x}<\mathrm{1}<\frac{\pi}{\mathrm{2}}\:\Rightarrow\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)<\mathrm{cos}\:{x}\:\checkmark \\ $$
Commented by mr W last updated on 30/Oct/24
$$\left.{I}\:{use}\:{my}\:{brain}\:{only}\::\right) \\ $$
Commented by Frix last updated on 31/Oct/24
$$\mathrm{It}'\mathrm{s}\:\mathrm{good}\:\mathrm{you}\:\mathrm{use}\:{your}\:\mathrm{brain}\:\mathrm{only}\:\mathrm{because} \\ $$$$\mathrm{brain}\:\mathrm{rents}\:\mathrm{had}\:\mathrm{been}\:\mathrm{raised}\:\mathrm{lately}. \\ $$
Commented by mr W last updated on 31/Oct/24
$$\left.{i}'{ve}\:{got}\:{a}\:{flat}\:{rate}\::\right) \\ $$
Answered by MrGaster last updated on 30/Oct/24
![sin(cos x)<cos x⇔sin(cos x)−cos x<0⇔2 cos(((cos x+(π/2)−x)/2))sin(((cos x−((π/2)−x))/2))<0⇔2 cos((π/4)+((x−cos x)/2))sin(((cos x+x−(π/2))/2))<0 cos x<cos(sin x)⇔cos x−cos(sin x)<0⇔−2 sin(((x+sin x)/2))sin(((x−sin x)/2))<0⇔2 sin(((x+sin x)/2))sin(((sin x−x)/2))>0 [Q.E.D]](https://www.tinkutara.com/question/Q213095.png)
$$\mathrm{sin}\left(\mathrm{cos}\:{x}\right)<\mathrm{cos}\:{x}\Leftrightarrow\mathrm{sin}\left(\mathrm{cos}\:{x}\right)−\mathrm{cos}\:{x}<\mathrm{0}\Leftrightarrow\mathrm{2}\:\mathrm{cos}\left(\frac{\mathrm{cos}\:{x}+\frac{\pi}{\mathrm{2}}−{x}}{\mathrm{2}}\right)\mathrm{sin}\left(\frac{\mathrm{cos}\:{x}−\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{\mathrm{2}}\right)<\mathrm{0}\Leftrightarrow\mathrm{2}\:\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}−\mathrm{cos}\:{x}}{\mathrm{2}}\right)\mathrm{sin}\left(\frac{\mathrm{cos}\:{x}+{x}−\frac{\pi}{\mathrm{2}}}{\mathrm{2}}\right)<\mathrm{0} \\ $$$$\mathrm{cos}\:{x}<\mathrm{cos}\left(\mathrm{sin}\:{x}\right)\Leftrightarrow\mathrm{cos}\:{x}−\mathrm{cos}\left(\mathrm{sin}\:{x}\right)<\mathrm{0}\Leftrightarrow−\mathrm{2}\:\mathrm{sin}\left(\frac{{x}+\mathrm{sin}\:{x}}{\mathrm{2}}\right)\mathrm{sin}\left(\frac{{x}−\mathrm{sin}\:{x}}{\mathrm{2}}\right)<\mathrm{0}\Leftrightarrow\mathrm{2}\:\mathrm{sin}\left(\frac{{x}+\mathrm{sin}\:{x}}{\mathrm{2}}\right)\mathrm{sin}\left(\frac{\mathrm{sin}\:{x}−{x}}{\mathrm{2}}\right)>\mathrm{0} \\ $$$$\left[\mathbb{Q}.\mathbb{E}.\mathbb{D}\right] \\ $$
Answered by Berbere last updated on 31/Oct/24
$${sin}\left({x}\right)\leqslant{x} \\ $$$${cos}\left({x}\right)\leqslant\mathrm{1}\Rightarrow\int_{\mathrm{0}} ^{{x}} {cos}\left({t}\right)\leqslant\int_{\mathrm{0}} ^{{x}} \mathrm{1}{dt}\Rightarrow{sin}\left({x}\right)\leqslant{x}\Rightarrow{sin}\left({cos}\left({x}\right)\right)\leqslant{cos}\left({x}\right) \\ $$$${cos}\left({sin}\left({x}\right)\right)>{cos}\left({x}\right) \\ $$$${sin}\left({x}\right)<{x}\:{by}\:{provious}\: \\ $$$${cos}\:{decrease}\Rightarrow{cos}\left({sin}\left({x}\right)\right)\geqslant{cos}\left({x}\right) \\ $$$$ \\ $$