Two Functions S and C Are Given

Let S and C be two functions. Assume that the domain for both S and C is the set of all real numbers and that S and C satisfy the following two identities.

S(x - y) = S(x)*C(y) - C(x)*S(y)
AND
C(x - y) = C(x)*C(y) + S(x)*S(y)

Also, suppose that the function S is NOT identically zero. That is,
S(x) DOES NOT equal zero for at least one real number x.

(1) SHOW THAT S(0) = 0.
(2) SHOW THAT C(0) = 1.
(3) PROVE THE IDENTITY [C(x)]^2 + [S(x)]^2 = 1.

(1) If the two identities are valid for all real numbers then they are valid for x = 0 and y=0
S(0)=S(0)*C(0)-C(0)*S(0)=0
(2) C(0)=C(0)*C(0) + S(0)*S(0)
C(0)=[C(0)]^2
C(0)=0 or C(0)=1
If C(0)=0 then S(x)=S(x-0)=S(x)*C(0)-C(x)*S(0)=0
Since S is NOT identically zero, C(0)=1
(3) Since C(0) = 1 and C(x - y) = C(x)*C(y) + S(x)*S(y),
1 = C(0)=C(x-x)=C(x)*C(x)+S(x)*S(x)=[C(x)]^2 + [S(x)]^2

1)
S(0) =
S(x - x) =
S(x)*C(x) - C(x)*S(x) =
0

2)
S(x) =
S(x - 0) =
S(x)C(0) - C(x)S(0) = ... since S(0) = 0
S(x)*C(0)

Thus, S(x) = S(x)*C(0). Since S(x) is not identically zero, we may say that C(0) = S(x)/S(x) = 1.

3)
1 =
C(0) =
C(x - x) =
C(x)*C(x) + S(x)*S(x) =
[C(x)]^2 + [S(x)]^2

QED