y-arc-tan-1-cosx-1-cosx-y-

Question Number 64514 by Mikael last updated on 18/Jul/19

$${y}={arc}\:{tan}\left[\sqrt{\frac{\mathrm{1}−{cosx}}{\mathrm{1}+{cosx}}}\right] \\ $$$${y}^{} =? \\ $$

Commented by mathmax by abdo last updated on 18/Jul/19

$${we}\:{have}\:\frac{\mathrm{1}−{cosx}}{\mathrm{1}+{cosx}}\:=\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}\:={tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\:\Rightarrow\sqrt{\frac{\mathrm{1}−{cosx}}{\mathrm{1}+{cosx}}}=\mid{tan}\left(\frac{{x}}{\mathrm{2}}\right)\mid \\ $$$${case}\:\mathrm{1}\:\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)>\mathrm{0}\:\Rightarrow{y}\left({x}\right)=\frac{{x}}{\mathrm{2}}\:\Rightarrow{y}^{'} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${case}\:\mathrm{2}\:\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)<\mathrm{0}\:\Rightarrow{y}\left({x}\right)\:=−\frac{{x}}{\mathrm{2}}\:\Rightarrow{y}^{'} \left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by Mikael last updated on 19/Jul/19

$${thank}\:{you}\:{Sir} \\ $$

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