Question Number 34845 by candre last updated on 11/May/18
![f(x)=cos(x) g(x)=2^x f:R→R g:R→R^+ [f(x)]^2 +[f(π/2−x)]^2 =((log_2 g(x))/x);x≠0 f(g(x)x)=[f(g(x−1)x)]^2 +[f(π/2−g(x−1)x)]^2 find f and g](https://www.tinkutara.com/question/Q34845.png)
$${f}\left({x}\right)=\mathrm{cos}\left({x}\right) \\ $$$${g}\left({x}\right)=\mathrm{2}^{{x}} \\ $$$${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R}^{+} \\ $$$$\left[{f}\left({x}\right)\right]^{\mathrm{2}} +\left[{f}\left(\pi/\mathrm{2}−{x}\right)\right]^{\mathrm{2}} =\frac{\mathrm{log}_{\mathrm{2}} {g}\left({x}\right)}{{x}};{x}\neq\mathrm{0} \\ $$$${f}\left({g}\left({x}\right){x}\right)=\left[{f}\left({g}\left({x}−\mathrm{1}\right){x}\right)\right]^{\mathrm{2}} +\left[{f}\left(\pi/\mathrm{2}−{g}\left({x}−\mathrm{1}\right){x}\right)\right]^{\mathrm{2}} \\ $$$$\mathrm{find}\:{f}\:\mathrm{and}\:{g} \\ $$