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Question Number 30528 by abdo imad last updated on 22/Feb/18

simplify A= arctan(((sinx)/(1−cosx))) .

$${simplify}\:\:{A}=\:{arctan}\left(\frac{{sinx}}{\mathrm{1}−{cosx}}\right)\:. \\ $$

Commented by abdo imad last updated on 23/Feb/18

we have A=arctan(((2sin((x/2))cos((x/2)))/(2sin^2 ((x/2))))) A=arctan(cotan((x/2)))=arctan( (1/(tan((x/2))))) case 1 if 0<x<π tan((x/2))>0⇒A=(π/2) −arctan(tan((x/2))) ⇒ A=(π/2) −(x/2)=((π−x)/2) case2 if −π<x<0 tan((x/2))<0 ⇒A=−(π/2) −arctan(tan((x/2))) ⇒A=−(π/2) −(x/2) .

$${we}\:{have}\:{A}={arctan}\left(\frac{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}\right) \\ $$$${A}={arctan}\left({cotan}\left(\frac{{x}}{\mathrm{2}}\right)\right)={arctan}\left(\:\frac{\mathrm{1}}{{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\right) \\ $$$${case}\:\mathrm{1}\:{if}\:\mathrm{0}<{x}<\pi\:\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)>\mathrm{0}\Rightarrow{A}=\frac{\pi}{\mathrm{2}}\:−{arctan}\left({tan}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$\Rightarrow\:{A}=\frac{\pi}{\mathrm{2}}\:−\frac{{x}}{\mathrm{2}}=\frac{\pi−{x}}{\mathrm{2}} \\ $$$${case}\mathrm{2}\:{if}\:−\pi<{x}<\mathrm{0}\:\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)<\mathrm{0}\:\Rightarrow{A}=−\frac{\pi}{\mathrm{2}}\:−{arctan}\left({tan}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$\Rightarrow{A}=−\frac{\pi}{\mathrm{2}}\:−\frac{{x}}{\mathrm{2}}\:. \\ $$

Answered by prakash jain last updated on 22/Feb/18

((sin x)/(1−cos x))=((2sin (x/2)cos (x/2))/(2sin^2 (x/2)))=cot (x/2) arctan(((sinx)/(1−cosx))) =tan^(−1) cot (x/2)

$$\frac{\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{cos}\:{x}}=\frac{\mathrm{2sin}\:\frac{{x}}{\mathrm{2}}\mathrm{cos}\:\frac{{x}}{\mathrm{2}}}{\mathrm{2sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}=\mathrm{cot}\:\frac{{x}}{\mathrm{2}} \\ $$$${arctan}\left(\frac{{sinx}}{\mathrm{1}−{cosx}}\right)\:=\mathrm{tan}^{−\mathrm{1}} \mathrm{cot}\:\frac{{x}}{\mathrm{2}} \\ $$

Commented by prakash jain last updated on 23/Feb/18

what if x=8π, 9π etc? tan^(−1) x should be from −(π/2) to (π/2)

$$\mathrm{what}\:\mathrm{if}\:{x}=\mathrm{8}\pi,\:\mathrm{9}\pi\:\mathrm{etc}? \\ $$$$\mathrm{tan}^{−\mathrm{1}} {x}\:\mathrm{should}\:\mathrm{be}\:\mathrm{from}\:−\frac{\pi}{\mathrm{2}}\:\mathrm{to}\:\frac{\pi}{\mathrm{2}} \\ $$

Commented by mrW2 last updated on 23/Feb/18

tan^(−1) cot (x/2)=(π/2)−(x/2) ???

$$\mathrm{tan}^{−\mathrm{1}} \mathrm{cot}\:\frac{{x}}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\:??? \\ $$

Commented by mrW2 last updated on 23/Feb/18

for 0<x<2π: A=(1/2)(π−x) generally: A=(1/2)(π+2π[(x/(2π))]−x) e.g.: x=6.5π [(x/(2π))]=[((6.5π)/(2π))]=3 A=(1/2)(π+6π−6.5π)=(π/4)

$${for}\:\mathrm{0}<{x}<\mathrm{2}\pi: \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi−{x}\right) \\ $$$$ \\ $$$${generally}: \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi+\mathrm{2}\pi\left[\frac{{x}}{\mathrm{2}\pi}\right]−{x}\right) \\ $$$$ \\ $$$${e}.{g}.:\: \\ $$$${x}=\mathrm{6}.\mathrm{5}\pi \\ $$$$\left[\frac{{x}}{\mathrm{2}\pi}\right]=\left[\frac{\mathrm{6}.\mathrm{5}\pi}{\mathrm{2}\pi}\right]=\mathrm{3} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi+\mathrm{6}\pi−\mathrm{6}.\mathrm{5}\pi\right)=\frac{\pi}{\mathrm{4}} \\ $$

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