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Question Number 117568 by ZiYangLee last updated on 12/Oct/20

$$\mathrm{Suppose}\mathrm{the}\mathrm{non}-\mathrm{constant}\mathrm{functions}f\mathrm{and}g$$$$\mathrm{satisfy}\mathrm{the}\mathrm{following}\mathrm{two}\mathrm{conditions}:$$$$\mathrm{I}:g(x-y)=g\left(x\right)g\left(y\right)+f\left(x\right)f\left(y\right)\mathrm{\forall }x,y\in \mathbb{R}$$$$\mathrm{II}:f\left(0\right)=0$$$$\mathrm{Evaluate}$$$$\mathrm{i}.g\left(0\right)\mathrm{ii}.{\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2$$

Commented by prakash jain last updated on 13/Oct/20

$$g\left(x\right)=\cos \left(x\right)$$$$f\left(x\right)=\sin \left(x\right)$$

Answered by Olaf last updated on 12/Oct/20

$$$$$$\mathrm{If}x=y$$$$g\left(0\right)={\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2\left(1\right)$$$$$$$$\mathrm{If}x=y=0$$$$g\left(0\right)={\left[f\left(0\right)\right]}^2+{\left[g\left(0\right)\right]}^2={\left[g\left(0\right)\right]}^2$$$$\Rightarrow g\left(0\right)=0\mathrm{or}g\left(0\right)=1$$$$$$$$\mathrm{But}\mathrm{if}g\left(0\right)=0,\left(1\right)\Rightarrow f\left(x\right)=g\left(x\right)=0$$$$\mathrm{Then}f\mathrm{and}g\mathrm{are}\mathrm{constant}\mathrm{functions}.$$$$\mathrm{Impossible}!$$$$$$$$\mathrm{Finally}g\left(0\right)=1\mathrm{and}:$$$${\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2=1$$

Commented by ZiYangLee last updated on 12/Oct/20

$$\mathrm{Good}!!$$

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