Question Number 145938 by Mathspace last updated on 09/Jul/21
$${g}\left({x}\right)={cos}\left({arctanx}\right) \\ $$$${if}\:{g}\left({x}\right)=\Sigma\:{a}_{{n}} {x}^{{n}} \:{determine}\:{the} \\ $$$${sequence}\:{a}_{{n}} \\ $$
Answered by Olaf_Thorendsen last updated on 09/Jul/21
![If x∈R, artanx ∈]−(π/2),+(π/2)[ ⇒ cos(arctanx) > 0 cos(arctanx) = +(√(cos^2 (arctanx))) cos(arctanx) = (√((cos^2 (arctanx))/(cos^2 (actanx)+sin^2 (arctanx)))) cos(arctanx) = (1/( (√(1+tan^2 (arctanx))))) = (1/( (√(1+x^2 )))) a_(2n) = (−1)^n ((1.3.5...(2n−1))/(2.4.6...(2n))) a_(2n) = (−1)^n (((2n)!)/(2^(2n) n!^2 )) and a_(2n+1) = 0](https://www.tinkutara.com/question/Q145945.png)
$$\left.\mathrm{If}\:{x}\in\mathbb{R},\:\mathrm{artan}{x}\:\in\right]−\frac{\pi}{\mathrm{2}},+\frac{\pi}{\mathrm{2}}\left[\right. \\ $$$$\Rightarrow\:\mathrm{cos}\left(\mathrm{arctan}{x}\right)\:>\:\mathrm{0} \\ $$$$\mathrm{cos}\left(\mathrm{arctan}{x}\right)\:=\:+\sqrt{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{arctan}{x}\right)} \\ $$$$\mathrm{cos}\left(\mathrm{arctan}{x}\right)\:=\:\sqrt{\frac{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{arctan}{x}\right)}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{actan}{x}\right)+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{arctan}{x}\right)}} \\ $$$$\mathrm{cos}\left(\mathrm{arctan}{x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{arctan}{x}\right)}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${a}_{\mathrm{2}{n}} \:=\:\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{1}.\mathrm{3}.\mathrm{5}…\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{2}.\mathrm{4}.\mathrm{6}…\left(\mathrm{2}{n}\right)} \\ $$$${a}_{\mathrm{2}{n}} \:=\:\left(−\mathrm{1}\right)^{{n}} \frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} {n}!^{\mathrm{2}} } \\ $$$$\mathrm{and}\:{a}_{\mathrm{2}{n}+\mathrm{1}} \:=\:\mathrm{0} \\ $$