Question Number 117568 by ZiYangLee last updated on 12/Oct/20
![Suppose the non-constant functions f and g satisfy the following two conditions: I: g(x−y)=g(x)g(y)+f(x)f(y) ∀ x,y∈R II: f(0)=0 Evaluate i. g(0) ii.[f(x)]^2 +[g(x)]^2](https://www.tinkutara.com/question/Q117568.png)
$$\mathrm{Suppose}\:\mathrm{the}\:\mathrm{non}-\mathrm{constant}\:\mathrm{functions}\:{f}\:\mathrm{and}\:{g} \\ $$$$\mathrm{satisfy}\:\mathrm{the}\:\mathrm{following}\:\mathrm{two}\:\mathrm{conditions}: \\ $$$$\mathrm{I}:\:{g}\left({x}−{y}\right)={g}\left({x}\right){g}\left({y}\right)+{f}\left({x}\right){f}\left({y}\right)\:\forall\:{x},{y}\in\mathbb{R} \\ $$$$\mathrm{II}:\:{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{Evaluate} \\ $$$$\mathrm{i}.\:{g}\left(\mathrm{0}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{ii}.\left[{f}\left({x}\right)\right]^{\mathrm{2}} +\left[{g}\left({x}\right)\right]^{\mathrm{2}} \\ $$
Commented by prakash jain last updated on 13/Oct/20
$${g}\left({x}\right)=\mathrm{cos}\:\left({x}\right) \\ $$$${f}\left({x}\right)=\mathrm{sin}\:\left({x}\right) \\ $$
Answered by Olaf last updated on 12/Oct/20
![If x = y g(0) = [f(x)]^2 +[g(x)]^2 (1) If x = y = 0 g(0) = [f(0)]^2 +[g(0)]^2 = [g(0)]^2 ⇒ g(0) = 0 or g(0) = 1 But if g(0) = 0, (1) ⇒ f(x) = g(x) = 0 Then f and g are constant functions. Impossible ! Finally g(0) = 1 and : [f(x)]^2 +[g(x)]^2 = 1](https://www.tinkutara.com/question/Q117573.png)
$$ \\ $$$$\mathrm{If}\:{x}\:=\:{y} \\ $$$${g}\left(\mathrm{0}\right)\:=\:\left[{f}\left({x}\right)\right]^{\mathrm{2}} +\left[{g}\left({x}\right)\right]^{\mathrm{2}} \:\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{If}\:{x}\:=\:{y}\:=\:\mathrm{0} \\ $$$${g}\left(\mathrm{0}\right)\:=\:\left[{f}\left(\mathrm{0}\right)\right]^{\mathrm{2}} +\left[{g}\left(\mathrm{0}\right)\right]^{\mathrm{2}} \:=\:\left[{g}\left(\mathrm{0}\right)\right]^{\mathrm{2}} \\ $$$$\Rightarrow\:{g}\left(\mathrm{0}\right)\:=\:\mathrm{0}\:\mathrm{or}\:{g}\left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$$$ \\ $$$$\mathrm{But}\:\mathrm{if}\:{g}\left(\mathrm{0}\right)\:=\:\mathrm{0},\:\left(\mathrm{1}\right)\:\Rightarrow\:{f}\left({x}\right)\:=\:{g}\left({x}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{Then}\:{f}\:\mathrm{and}\:{g}\:\mathrm{are}\:\mathrm{constant}\:\mathrm{functions}. \\ $$$$\mathrm{Impossible}\:! \\ $$$$ \\ $$$$\mathrm{Finally}\:{g}\left(\mathrm{0}\right)\:=\:\mathrm{1}\:\mathrm{and}\:: \\ $$$$\left[{f}\left({x}\right)\right]^{\mathrm{2}} +\left[{g}\left({x}\right)\right]^{\mathrm{2}} \:=\:\mathrm{1} \\ $$
Commented by ZiYangLee last updated on 12/Oct/20
$$\mathrm{Good}!! \\ $$