Question Number 179253 by mathlove last updated on 27/Oct/22
$$\left({fog}\right)_{{x}} ={cos}\mathrm{2}{x}\:\:{and}\:{g}\left({x}\right)={tanx} \\ $$$${f}\left({x}\right)=? \\ $$
Commented by CElcedricjunior last updated on 30/Oct/22
![(fog)(x)=cos2x ou^ g(x)=tanx Determinons f(x) (fog)(x)=cos2x =>f[g(x)]=cos2x =>f(tanx)=cos2x En composant x pas arctanx on a tan(arctanx)=x =>f(tan(arctanx))=cos[2(arctanx)] =>f(x)=cos[2(arctanx)] ..............lecelebre cedric junior.......](https://www.tinkutara.com/question/Q179561.png)
$$\left({fog}\right)\left({x}\right)=\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}\:\boldsymbol{{o}}\grave {\boldsymbol{{u}}}\:\boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{x}}\right)=\boldsymbol{\mathrm{tanx}} \\ $$$$\boldsymbol{\mathcal{D}{eterminons}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right) \\ $$$$\left(\boldsymbol{{fog}}\right)\left(\boldsymbol{{x}}\right)=\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}} \\ $$$$=>\boldsymbol{{f}}\left[\boldsymbol{{g}}\left(\boldsymbol{{x}}\right)\right]=\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}} \\ $$$$=>\boldsymbol{{f}}\left(\boldsymbol{{tanx}}\right)=\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}} \\ $$$$\mathfrak{E}\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{composant}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{pas}}\:\boldsymbol{\mathrm{arctanx}} \\ $$$$\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{tan}}\left(\boldsymbol{\mathrm{arctanx}}\right)=\boldsymbol{\mathrm{x}} \\ $$$$=>\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{tan}}\left(\boldsymbol{\mathrm{arctanx}}\right)\right)=\boldsymbol{\mathrm{cos}}\left[\mathrm{2}\left(\boldsymbol{\mathrm{arctanx}}\right)\right] \\ $$$$=>\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\boldsymbol{\mathrm{cos}}\left[\mathrm{2}\left(\boldsymbol{\mathrm{arctanx}}\right)\right] \\ $$$$…………..{lecelebre}\:{cedric}\:{junior}……. \\ $$
Answered by mr W last updated on 27/Oct/22
$${f}\left(\mathrm{tan}\:{x}\right)=\mathrm{cos}\:\mathrm{2}{x} \\ $$$${t}=\mathrm{tan}\:{x}\:\Rightarrow{x}=\mathrm{tan}^{−\mathrm{1}} {t} \\ $$$$\Rightarrow{f}\left({t}\right)=\mathrm{cos}\:\left(\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} {t}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{cos}\:\left(\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} {x}\right) \\ $$
Commented by mathlove last updated on 27/Oct/22
$${thanks}\:{sir}\:{mr}\:{W} \\ $$
Commented by MJS_new last updated on 27/Oct/22
$$\mathrm{cos}\:\mathrm{2arctan}\:{x}\:=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$
Commented by mathlove last updated on 27/Oct/22
$${HOW}\:{is}\:{this}? \\ $$
Commented by mr W last updated on 27/Oct/22
$${MJS}\:{sir}\:{is}\:{right}! \\ $$$$\mathrm{cos}\:\mathrm{2}{t}=\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:{t}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{t}} \\ $$