functional equation (2)

[man111]

if [tex]f(x+y)=f(x)g(y)+g(x)f(y)[/tex] and [tex]g(x+y)=g(x)g(y)-f(x)f(y)[/tex].Then

prove that [tex]f(0)=0[/tex] and [tex]g(0)=1[/tex]

*If There is any Trigonometric Function That satisfy above functional equation.
If Yes,Then How Can I calculate that.

[mkmarinov]

Obviously f(x)=sinx and g(x)=cosx are the trigonometric functions, satisfying the equations.
Plugging in x=y=0, assuming that g(0) != 0:
[tex]f(0)=2f(0)g(0) => f(0)(g(0)-\frac{1}{2})=0[/tex] (1)
[tex]\frac{f(0)}{g(0)}=2f(0)[/tex] (2)

In the second equation:
[tex]g(0)=g^2(0)-f^2(0) / :g(0)[/tex]
[tex]1=g(0)-2f^2(0)[/tex]
[tex]g(0)=1+2f^2(0) \ge 1[/tex]

But from (1) we get either f(0)=0 or g(0)=1/2. The second is impossible, so f(0)=0 and g(0)=1.

But g(0)=f(0)=0 is also a solution. Then f(x)=g(x)=0 for any x.

I did not understand if it must be proven that f(x)=sinx and g(x)=cosx, or if you just needed to plug in these functions to show that they satisfy the given equation.

[man111]

Thanks mkmarinov.

Here my second Question is find all function [tex]f(x)[/tex] and [tex]g(x)[/tex]that satisfy given functional equation.

[mkmarinov]

Well, you have to tell us (at least) where they are defined. And also if they should be continuous, differentiable, etc.