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Question Number 192758 by vishal1234 last updated on 26/May/23
what is the value of dy/dx if y = cot^(−1) [(((√(1+sinx))+(√(1−sinx)))/( (√(1+sinx))−(√(1−sinx))))]; x∈(0,(π/2)) please give complete explanation that why can we not take [sinx/2−cosx/2]^2 = 1−sinx
$${what}\:{is}\:{the}\:{value}\:{of}\:{dy}/{dx}\:{if} \\ $$$${y}\:=\:{cot}^{−\mathrm{1}} \left[\frac{\sqrt{\mathrm{1}+{sinx}}+\sqrt{\mathrm{1}−{sinx}}}{\:\sqrt{\mathrm{1}+{sinx}}−\sqrt{\mathrm{1}−{sinx}}}\right];\:{x}\in\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$$${please}\:{give}\:{complete}\:{explanation}\:{that} \\ $$$${why}\:{can}\:{we}\:{not}\:{take} \\ $$$$\left[{sinx}/\mathrm{2}−{cosx}/\mathrm{2}\right]^{\mathrm{2}} \:=\:\mathrm{1}−{sinx} \\ $$
Answered by a.lgnaoui last updated on 26/May/23
for the second question (sin (x/2)−cos (x/2) )^2 =(sin (x/2))^2 +(cos (x/2))^2 −2(sin (x/2))(cos (x/2)) = 1±2[(sin (x/2))(√(1−(sin (x/2))^2 )) posons u=sin (x/2) z=(sin (x/2)−cos (x/2))^2 { ((z=1+2u(√(1−u^2 )) (1))),((z=1−2u(√(1−u^2 )) (2))) :} 1+2u(√(1−u^2 )) =1+2sin (x/2)cos (x/2) =1+sin x 1−2u(√(1−u^2 )) =1−sin x ⇒(sin (x/2)−cos (x/2))^2 =1+∣sin x∣ •−(π/2)+2kπ<x<+(π/2)+2kπ sin x>0 z=1+sin x •(π/2)+2kπ<x<((3π)/2)+2kπ sin x<0 z=1−sin x donc z n est pas toujours egale ( 1−sinx)
$$\mathrm{for}\:\mathrm{the}\:\mathrm{second}\:\mathrm{question} \\ $$$$\left(\mathrm{sin}\:\frac{\mathrm{x}}{\mathrm{2}}−\mathrm{cos}\:\frac{\mathrm{x}}{\mathrm{2}}\:\right)^{\mathrm{2}} =\left(\mathrm{sin}\:\frac{\mathrm{x}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{cos}\:\frac{\mathrm{x}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\left(\mathrm{sin}\:\frac{\mathrm{x}}{\mathrm{2}}\right)\left(\mathrm{cos}\:\frac{\mathrm{x}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{1}\pm\mathrm{2}\left[\left(\mathrm{sin}\:\frac{\mathrm{x}}{\mathrm{2}}\right)\sqrt{\mathrm{1}−\left(\mathrm{sin}\:\frac{\mathrm{x}}{\mathrm{2}}\right)^{\mathrm{2}} }\right. \\ $$$$\:\:\:\mathrm{posons}\:\:\:\:\mathrm{u}=\mathrm{sin}\:\frac{\mathrm{x}}{\mathrm{2}}\:\:\:\:\mathrm{z}=\left(\mathrm{sin}\:\frac{\mathrm{x}}{\mathrm{2}}−\mathrm{cos}\:\frac{\mathrm{x}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\begin{cases}{\mathrm{z}=\mathrm{1}+\mathrm{2u}\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} \:}\:\:\:\:\:\:\:\left(\mathrm{1}\right)}\\{\mathrm{z}=\mathrm{1}−\mathrm{2u}\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\left(\mathrm{2}\right)}\end{cases}\:\:\:\:\:\:\:\: \\ $$$$\mathrm{1}+\mathrm{2u}\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }\:\:\:=\mathrm{1}+\mathrm{2sin}\:\frac{\mathrm{x}}{\mathrm{2}}\mathrm{cos}\:\frac{\mathrm{x}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\mathrm{sin}\:\mathrm{x} \\ $$$$\mathrm{1}−\mathrm{2u}\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }\:\:=\mathrm{1}−\mathrm{sin}\:\mathrm{x} \\ $$$$ \\ $$$$\Rightarrow\left(\mathrm{sin}\:\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}−\mathrm{cos}\:\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{1}+\mid\mathrm{sin}\:\boldsymbol{\mathrm{x}}\mid \\ $$$$ \\ $$$$\:\:\:\bullet−\frac{\pi}{\mathrm{2}}+\mathrm{2k}\pi<\mathrm{x}<+\frac{\pi}{\mathrm{2}}+\mathrm{2k}\pi \\ $$$$\:\:\:\mathrm{sin}\:\mathrm{x}>\mathrm{0}\:\:\:\:\:\mathrm{z}=\mathrm{1}+\mathrm{sin}\:\mathrm{x} \\ $$$$\:\:\:\:\bullet\frac{\pi}{\mathrm{2}}+\mathrm{2k}\pi<\mathrm{x}<\frac{\mathrm{3}\pi}{\mathrm{2}}+\mathrm{2k}\pi \\ $$$$\:\:\:\:\mathrm{sin}\:\mathrm{x}<\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{z}=\mathrm{1}−\mathrm{sin}\:\mathrm{x} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{donc}}\:\:\:\boldsymbol{\mathrm{z}}\:\:\mathrm{n}\:\mathrm{est}\:\:\mathrm{pas}\:\boldsymbol{\mathrm{toujours}}\:\:\boldsymbol{\mathrm{egale}}\:\:\left(\:\mathrm{1}−\boldsymbol{\mathrm{sinx}}\right) \\ $$
Commented by a.lgnaoui last updated on 26/May/23
−(π/4)< (x/2)<(π/4) cas 1 (π/4)<(x/2)<π+(π/4) cas 2 • sin (x/2)cos (x/2)>0 cas1 ⇒z=1−sin x •sin (x/2)cos (x/2)<0 cas 2 ⇒ z=1+sin x
$$\:\:\:\:\:−\frac{\pi}{\mathrm{4}}<\:\frac{\mathrm{x}}{\mathrm{2}}<\frac{\pi}{\mathrm{4}}\:\:\:\:\:\:\:\:\mathrm{cas}\:\mathrm{1}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\pi}{\mathrm{4}}<\frac{\mathrm{x}}{\mathrm{2}}<\pi+\frac{\pi}{\mathrm{4}}\:\:\:\mathrm{cas}\:\mathrm{2} \\ $$$$\:\:\:\bullet\:\mathrm{sin}\:\frac{\mathrm{x}}{\mathrm{2}}\mathrm{cos}\:\frac{\mathrm{x}}{\mathrm{2}}>\mathrm{0}\:\:\:\:\mathrm{cas1}\:\:\:\Rightarrow\mathrm{z}=\mathrm{1}−\mathrm{sin}\:\mathrm{x} \\ $$$$\:\:\:\:\bullet\mathrm{sin}\:\frac{\mathrm{x}}{\mathrm{2}}\mathrm{cos}\:\frac{\mathrm{x}}{\mathrm{2}}<\mathrm{0}\:\:\:\:\:\mathrm{cas}\:\mathrm{2}\:\:\Rightarrow\:\:\mathrm{z}=\mathrm{1}+\mathrm{sin}\:\mathrm{x} \\ $$
Answered by mahdipoor last updated on 26/May/23
why can not take ?! take that, bot pay attention : (√(1−sinx))=(√((sin(x/2)−cos(x/2))^2 ))= ∣sin(x/2)−cos(x/2)∣=cos(x/2)−sin(x/2) at 0<x<(π/2) and (√(1+sinx))=cos(x/2)+sin(x/2) y=cot^(−1) (((cos(x/2))/(sin(x/2))))=cot^(−1) (cot((x/2)))=(x/2)⇒(dy/dx)=(1/2)
$${why}\:{can}\:{not}\:{take}\:?!\: \\ $$$${take}\:{that},\:\:{bot}\:{pay}\:{attention}\:: \\ $$$$\sqrt{\mathrm{1}−{sinx}}=\sqrt{\left({sin}\frac{{x}}{\mathrm{2}}−{cos}\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} }= \\ $$$$\mid{sin}\frac{{x}}{\mathrm{2}}−{cos}\frac{{x}}{\mathrm{2}}\mid={cos}\frac{{x}}{\mathrm{2}}−{sin}\frac{{x}}{\mathrm{2}}\:\:\:{at}\:\mathrm{0}<{x}<\frac{\pi}{\mathrm{2}} \\ $$$${and}\: \\ $$$$\sqrt{\mathrm{1}+{sinx}}={cos}\frac{{x}}{\mathrm{2}}+{sin}\frac{{x}}{\mathrm{2}} \\ $$$${y}={cot}^{−\mathrm{1}} \left(\frac{{cos}\frac{{x}}{\mathrm{2}}}{{sin}\frac{{x}}{\mathrm{2}}}\right)={cot}^{−\mathrm{1}} \left({cot}\left(\frac{{x}}{\mathrm{2}}\right)\right)=\frac{{x}}{\mathrm{2}}\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by a.lgnaoui last updated on 26/May/23
First question cot y=(((√(1+sin x)) +(√(1−sin x)))/( (√(1+sin x)) −(√(1−sin x)))) =(([(√(1+sin x)) +(√(1−sin x)) ]^2 )/(2sin x)) =((1+∣cos x∣)/(sin x)) 0 <x<(π/2) ⇒ cos x>0 ⇒ cot y=((1+cos x)/(sin x)) ((d(cot y))/dx)=((d(cot y))/dy)×(dy/dx) −((cos x+1)/(sin^2 x))=−(1/(sin^2 y))×(dy/dx) ((1+cos x)/(sin^2 x))sin^2 y =(dy/(dx )) (dy/dx)=[ ((1+cos x)/(sin^2 x))](1+cot^2 y) =((1+cos x)/(sin^2 x ))(1+((1+cos x)/(sin x))) =((1+cos x)/(sin^2 x))+(((1+cos x)^2 )/(sin^3 x)) (dy/dx)=(((1+cos x)(sin x+cos x+1))/(sin^3 x))
$$\boldsymbol{\mathrm{First}}\:\:\boldsymbol{\mathrm{question}} \\ $$$$\:\: \\ $$$$\:\:\mathrm{cot}\:\boldsymbol{\mathrm{y}}=\frac{\sqrt{\mathrm{1}+\mathrm{sin}\:\boldsymbol{\mathrm{x}}}\:+\sqrt{\mathrm{1}−\mathrm{sin}\:\boldsymbol{\mathrm{x}}}}{\:\sqrt{\mathrm{1}+\mathrm{sin}\:\boldsymbol{\mathrm{x}}}\:−\sqrt{\mathrm{1}−\mathrm{sin}\:\boldsymbol{\mathrm{x}}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\left[\sqrt{\mathrm{1}+\mathrm{sin}\:\boldsymbol{\mathrm{x}}}\:+\sqrt{\mathrm{1}−\mathrm{sin}\:\boldsymbol{\mathrm{x}}}\:\right]^{\mathrm{2}} }{\mathrm{2sin}\:\boldsymbol{\mathrm{x}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}+\mid\mathrm{cos}\:\mathrm{x}\mid}{\mathrm{sin}\:\mathrm{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{0}\:<\mathrm{x}<\frac{\pi}{\mathrm{2}}\:\:\:\:\Rightarrow\:\mathrm{cos}\:\mathrm{x}>\mathrm{0} \\ $$$$\:\Rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cot}\:\boldsymbol{\mathrm{y}}=\frac{\mathrm{1}+\mathrm{cos}\:\boldsymbol{\mathrm{x}}}{\mathrm{sin}\:\boldsymbol{\mathrm{x}}} \\ $$$$ \\ $$$$\:\:\:\frac{\boldsymbol{\mathrm{d}}\left(\mathrm{cot}\:\boldsymbol{\mathrm{y}}\right)}{\boldsymbol{\mathrm{dx}}}=\frac{\boldsymbol{\mathrm{d}}\left(\mathrm{cot}\:\boldsymbol{\mathrm{y}}\right)}{\boldsymbol{\mathrm{dy}}}×\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}} \\ $$$$\:\:\:\:−\frac{\mathrm{cos}\:\boldsymbol{\mathrm{x}}+\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\mathrm{x}}}=−\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\mathrm{y}}}×\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}} \\ $$$$\:\:\:\frac{\mathrm{1}+\mathrm{cos}\:\boldsymbol{\mathrm{x}}}{\mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\mathrm{x}}}\mathrm{sin}^{\mathrm{2}} \boldsymbol{\mathrm{y}}\:=\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}\:} \\ $$$$\:\:\:\:\:\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\left[\:\:\frac{\mathrm{1}+\mathrm{cos}\:\boldsymbol{\mathrm{x}}}{\mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\mathrm{x}}}\right]\left(\mathrm{1}+\mathrm{cot}\:^{\mathrm{2}} \boldsymbol{\mathrm{y}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{1}+\mathrm{cos}\:\boldsymbol{\mathrm{x}}}{\mathrm{sin}^{\mathrm{2}} \boldsymbol{\mathrm{x}}\:}\left(\mathrm{1}+\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{x}}\right) \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}+\frac{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }{\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}} \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{d}\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}}=\frac{\left(\mathrm{1}+\mathrm{cos}\:\boldsymbol{\mathrm{x}}\right)\left(\mathrm{sin}\:\boldsymbol{\mathrm{x}}+\mathrm{cos}\:\boldsymbol{\mathrm{x}}+\mathrm{1}\right)}{\mathrm{sin}\:^{\mathrm{3}} \boldsymbol{\mathrm{x}}} \\ $$$$ \\ $$

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