1@EmInEm
u cannot assume that!
if f(x+y)=f(x)g(y)+g(x)f(y)
and g(x+y)=g(x)g(y)-f(x)f(y)
prove that f(0)=0 and g(0)=1
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5 Answers
1obviously f = sinx , g = cos x but i kno u hav purposely put this question lol
341You cannot assume that.
f(x) = g(x) = 0 for all x is also a solution.
Assuming that is not the case, setting x=y=0, we have
f(0) = 2f(0) g(0) and g(0) = g2(0) - f2(0)
Either f(0) = 0 or g(0) = 1/2.
But if g(0) = 1/2, then Eqn 2 gives f2(0) = -1/2 which is not possible.
So, f(0) = 0.
From the second equation, g(0)(1-g(0)) = 0, so g(0) = 0 or g(0) =1.
Now, if g(0) =0, we have setting y=0 in the equations, f(x)=0 for all x and so g(x) for all x which we have already noted as a solution.
So we have the only other possibility as f(0) =0 and g(0) = 1
1yes i know that is not the only case but i thought that you wanted some one to come up with a proof that one solution is sinx, cosx .... just made a guess [6]