these identities quite literally define sin and cos

>>16871914
Implication =/= iff

Man, these new Captchas are really something else

Let f,g be differentiable and non-constant, and for all x,y:

f(x+y)=f(x)f(y)−g(x)g(y)
g(x+y)=f(x)g(y)+g(x)f(y)

1. Values at 0.
Put x=y=0:
f(0)=f(0)^2−g(0)^2
g(0)=2f(0)g(0)

If g(0)≠0 then 1=2f(0) so f(0)=1/2, and then f(0)=f(0)^2−g(0)^2 gives 1/2=1/4−g(0)^2, impossible.
So g(0)=0. Then f(0)=f(0)^2, so f(0)=0 or 1.
If f(0)=0, plug y=0 into the first equation: f(x)=f(x)f(0)=0 for all x, contradiction.
Hence f(0)=1 and g(0)=0.

2. Define H(x)=f(x)^2+g(x)^2.
Compute H(x+y):
H(x+y)=f(x+y)^2+g(x+y)^2
=(f(x)f(y)−g(x)g(y))^2+(f(x)g(y)+g(x)f(y))^2
= (f(x)^2+g(x)^2)(f(y)^2+g(y)^2)
=H(x)H(y)

Also H(0)=f(0)^2+g(0)^2=1.

3. Differentiate the multiplicative equation.
From H(x+y)=H(x)H(y), differentiate in y and set y=0:
H'(x)=H(x)H'(0)

So H solves the ODE H'(x)=cH(x) where c=H'(0), with H(0)=1, hence H(x)=e^{cx}.

4. Use f'(0)=0.
Compute H'(0):
H'(0)=2f(0)f'(0)+2g(0)g'(0)=2·1·0+0=0

So c=0, hence H(x)=e^{0}=1 for all x.

Therefore f(x)^2+g(x)^2=1 for all x.

>>16871921
Tell em unc

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>>16871921
You deserve tits

>>16871917
Who said it's an iff?

let h(x) = f(x)+ig(x)
=> h(x+y) = h(x)h(y)
=> h(x)=h(x)h(0)

f,g nonconstant
=> h(x) nonconstant
=> h(x) != 0 for some x

h(x) = h(x)h(0)
=> h(0) = 1
=> |h(0)| = f(0)^2+g(0)^2 = 1

diff the two given equations w.r.t. x then set x = 0 to get
f'(y) = -g'(0)g(y) = -Cg(y)
g'(y) = g'(0)f(y) = Cf(y)

f nonconstant so f' not identically zero so C != 0

d/dx (f^2+g^2) = 2ff'+2gg' = -2cfg+2cgf = 0

therefore f^2+g^2 is constant and = f(0)^2+g(0)^2 = 1

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>>16871909

>>16871909
f(2x) = f^2(x) - g^2(x)
g(2x) = 2f(x)g(x)
f^2(2x) = f^4(x) + g^4(x) -2f^2(x)g^2(x)
g^2(2x) = 4 f^2(x)g^(x)

f^2(2x) + g^2(2x) = [f^2(x) + g^2(x)]^2
Since this is true for all x, it must be the case that f^2(x) + g^2(x) = 1