Question-40276

Question Number 40276 by Raj Singh last updated on 18/Jul/18

Answered by MJS last updated on 18/Jul/18

(d/dx)[((sin x)/(1+tan x))]=((cos^3 x −sin^3 x)/(1+2sin x cos x)) zeros at x=(π/4)(4k+1) ∧ k∈Z max at x=(π/4)(4k+1) ∧ k=2n ∧ n∈Z ⇒ ⇒ max ay x=(π/4)(8n+1) ∧ n∈Z ∧ y=((√2)/4) min at x=(π/4)(1+4k) ∧ k=2n+1 ∧ n∈Z ⇒ ⇒ min at x=(π/4)(8n+5) ∧ n∈Z ∧ y=−((√2)/4)

$$\frac{{d}}{{dx}}\left[\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{tan}\:{x}}\right]=\frac{\mathrm{cos}^{\mathrm{3}} \:{x}\:−\mathrm{sin}^{\mathrm{3}} \:{x}}{\mathrm{1}+\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}} \\ $$$$\mathrm{zeros}\:\mathrm{at}\:{x}=\frac{\pi}{\mathrm{4}}\left(\mathrm{4}{k}+\mathrm{1}\right)\:\wedge\:{k}\in\mathbb{Z} \\ $$$$\mathrm{max}\:\mathrm{at}\:{x}=\frac{\pi}{\mathrm{4}}\left(\mathrm{4}{k}+\mathrm{1}\right)\:\wedge\:{k}=\mathrm{2}{n}\:\wedge\:{n}\in\mathbb{Z}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{max}\:\mathrm{ay}\:{x}=\frac{\pi}{\mathrm{4}}\left(\mathrm{8}{n}+\mathrm{1}\right)\:\wedge\:{n}\in\mathbb{Z}\:\wedge\:{y}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$\mathrm{min}\:\mathrm{at}\:{x}=\frac{\pi}{\mathrm{4}}\left(\mathrm{1}+\mathrm{4}{k}\right)\:\wedge\:{k}=\mathrm{2}{n}+\mathrm{1}\:\wedge\:{n}\in\mathbb{Z}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{min}\:\mathrm{at}\:{x}=\frac{\pi}{\mathrm{4}}\left(\mathrm{8}{n}+\mathrm{5}\right)\:\wedge\:{n}\in\mathbb{Z}\:\wedge\:{y}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jul/18

(dy/dx)=(((1+tanx)cosx−sinx(sec^2 x))/((1+tanx)^2 )) for max/min (dy/dx)=0 cosx+sinx−((sinx)/(cos^2 x))=0 cos^3 x+sinxcos^2 x−sinx=0 cos^3 x+sinx(1−sin^2 x)−sinx=0 cos^3 x−sin^3 x=0 tan^3 x=1 (tanx−1)(tan^2 x+tanx+1)=0 x=(Π/4) (dy/dx)=(((1+tanx)cosx−sinx(sec^2 x))/((1+tanx)^2 )) =((cosx+sinx−((sinx)/(cos^2 x)))/((1+tanx)^2 )) =((cos^3 x+sinxcos^2 x−sinx)/(cos^2 x(1+tanx)^2 )) =((cos^3 x+sinx(1−sin^2 x)−sinx)/(cos^2 x(1+tanx)^2 )) =((cos^3 x−sin^3 x)/(cos^2 x(1+tanx)^2 )) at x=(Π/4) (dy/dx)=0 cosx>sinx when x<(Π/4) (dy/dx)=+ve cosx<sinx when x>(Π/4) (dy/dx)=−ve so sign change of (dy/dx) from +ve to −ve so at x=(Π/4) ((sinx)/(1+tanx)) is maximum and max value is ((1/( (√2)))/(1+1))=(1/(2(√2) ))

$$\frac{{dy}}{{dx}}=\frac{\left(\mathrm{1}+{tanx}\right){cosx}−{sinx}\left({sec}^{\mathrm{2}} {x}\right)}{\left(\mathrm{1}+{tanx}\right)^{\mathrm{2}} } \\ $$$${for}\:{max}/{min}\:\:\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$${cosx}+{sinx}−\frac{{sinx}}{{cos}^{\mathrm{2}} {x}}=\mathrm{0} \\ $$$${cos}^{\mathrm{3}} {x}+{sinxcos}^{\mathrm{2}} {x}−{sinx}=\mathrm{0} \\ $$$${cos}^{\mathrm{3}} {x}+{sinx}\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)−{sinx}=\mathrm{0} \\ $$$${cos}^{\mathrm{3}} {x}−{sin}^{\mathrm{3}} {x}=\mathrm{0} \\ $$$${tan}^{\mathrm{3}} {x}=\mathrm{1} \\ $$$$\left({tanx}−\mathrm{1}\right)\left({tan}^{\mathrm{2}} {x}+{tanx}+\mathrm{1}\right)=\mathrm{0} \\ $$$$ \\ $$$${x}=\frac{\Pi}{\mathrm{4}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\left(\mathrm{1}+{tanx}\right){cosx}−{sinx}\left({sec}^{\mathrm{2}} {x}\right)}{\left(\mathrm{1}+{tanx}\right)^{\mathrm{2}} } \\ $$$$=\frac{{cosx}+{sinx}−\frac{{sinx}}{{cos}^{\mathrm{2}} {x}}}{\left(\mathrm{1}+{tanx}\right)^{\mathrm{2}} } \\ $$$$=\frac{{cos}^{\mathrm{3}} {x}+{sinxcos}^{\mathrm{2}} {x}−{sinx}}{{cos}^{\mathrm{2}} {x}\left(\mathrm{1}+{tanx}\right)^{\mathrm{2}} } \\ $$$$=\frac{{cos}^{\mathrm{3}} {x}+{sinx}\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)−{sinx}}{{cos}^{\mathrm{2}} {x}\left(\mathrm{1}+{tanx}\right)^{\mathrm{2}} } \\ $$$$=\frac{{cos}^{\mathrm{3}} {x}−{sin}^{\mathrm{3}} {x}}{{cos}^{\mathrm{2}} {x}\left(\mathrm{1}+{tanx}\right)^{\mathrm{2}} } \\ $$$${at}\:{x}=\frac{\Pi}{\mathrm{4}}\:\:\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$${cosx}>{sinx}\:\:{when}\:{x}<\frac{\Pi}{\mathrm{4}}\:\:\frac{{dy}}{{dx}}=+{ve} \\ $$$${cosx}<{sinx}\:\:{when}\:{x}>\frac{\Pi}{\mathrm{4}}\:\frac{{dy}}{{dx}}=−{ve} \\ $$$${so}\:{sign}\:{change}\:{of}\:\frac{{dy}}{{dx}}\:\:{from}\:+{ve}\:{to}\:−{ve}\: \\ $$$${so}\:{at}\:{x}=\frac{\Pi}{\mathrm{4}}\:\frac{{sinx}}{\mathrm{1}+{tanx}}\:{is}\:{maximum} \\ $$$${and}\:{max}\:{value}\:{is}\:\:\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{\mathrm{1}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\:} \\ $$$$ \\ $$

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