Prove $S^2(x) + C^2(x) = 1$ Suppose $S, C : \Bbb R \to \Bbb R$ are differentiable and that (1) $S'(x) = C(x)$ and $C'(x) = −S(x)$ for all $x \in \Bbb R$, and (2) $S(0) = 0$ and $C(0) = 1$. Prove that $S^2(x) + C^2(x) = 1$ for all $x \in \Bbb R$.
I tried using the limit definition of differentiable in setting the limit as $x$ approaches $a$ of $((S(X)-S(a))/x-1)=C(x)$ and setting the limit as $x$ approaches $a$ of $((C(X)-C(a))/x-1)=-S(x)$ but I don't know if that gets me anywhere. Hints or anything would be greatly appreciated!
 A: Note that
$\dfrac{d}{dx}(S^2(x) + C^2(x)) = 2S(x)S'(x) + 2C(x)C'(x) = 2(S(x)S'(x) + C(x)C'(x)); \tag 1$
by hypothesis
$S'(x) = C(x) \tag 2$
and
$C'(x) = -S(x); \tag 3$
thus (1) becomes
$\dfrac{d}{dx}(S^2(x) + C^2(x)) = 2(S(x)S'(x) + C(x)C'(x)) = 2(S(x)C(x) - C(x)S(x)) = 0; \tag 4$
with
$S(0) = 0, \tag 5$
and
$C(0) = 1, \tag 6$
we have
$S^2(0) + C^2(0) = 1; \tag 7$
by (4), $S^2(x) + C^2(x)$ is constant; thus from (7),
$\forall x \in \Bbb R, \; S^2(x) + C^2(x) = 1, \tag 8$
$OE\Delta$.
Note Added in Edit, Wedesday 4 September 2019 9:20 PM PST:  As pointed out in the comments to the question itself, $S(x)$ and $C(x)$ are in fact $\sin x$ and $\cos x$, respectively.  This may be seen as follows:  from (2), (3) we have
$S''(x) = C'(x) = -S(x), \tag 9$
and
$C''(x) = -S'(x) = -C(x); \tag{10}$
thus $S(x)$ and $C(x)$ each obey the differential equation
$y''(x) + y(x) = 0; \tag{11}$
it is well known that the unique non-vanishing solution to this equation with
$y(0) = 0, \; y'(0) = 1 \tag{12}$
is in fact $S(x)$; likewise $C(x)$ solves (11) with
$y(0) = 1, \; y'(0) = 0, \tag{13}$
where the initial conditions on $y'(0)$ are easily had from (2)-(3), (5)-(6).  Thus we infer that
$S(x) = \sin x, \; C(x) = \cos x. \tag{14}$
End of Note.
A: Given
$\frac{dS}{dx}=C(x) \Rightarrow SdS=S(x)C(x)dx \cdots (1)$
$\frac{dC}{dx}=-S(x) \Rightarrow CdC=-S(x)C(x)dx \cdots (2)$
Adding $(1)$ and $(2)$ and integrating, we have
$\int (SdS+CdC)=\int0dx \Rightarrow S^2(x)+C^2(x)=k \cdots (3)$
From the initial conditions of $S(0)$ & $C(0)$, we have $k=1$.
$\therefore$ From equation $(3)$, we have $S^2(x)+C^2(x)=1$.
