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This exercise can be found in the book by W. Trench: Introduction to Real Analysis: enter image description here

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    Here is possible start: first show that $S(0) = 0$ and $C(0) = 1$ then rewrite the first equation as $$\frac{C(x+y) - C(x)}{y} = C(x)\frac{C(y)-C(0)}{y} - S(x)\frac{S(y) - S(0)}{y}$$ and take $y\to 0$ and use the definition of the derivative to determine what each term approaches. This can be done for the second equation too. – 2017-02-01
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    Thank you so much - now I get it. It is not even so difficult as I had thought. – 2017-02-02

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Consider first $$ c(0)=c(0)c(0)-s(0)s(0), \qquad s(0)=s(0)c(0)+s(0)c(0) $$ by the main property with $x=y=0$. Try to deduce $c(0)=1$ and $s(0)=0$

From the second equation we conclude$$s(0)(1-2c(0))=0$$Thus either $s(0)=0$ or $c(0)=1/2$. If $c(0)=1/2$, the first equation gives$$\frac{1}{2}=\frac{1}{4}-(s(0))^2$$which is a contradiction. Hence $s(0)=0$ and $c(0)=(c(0))^2$, so either $c(0)=1$ or $c(0)=0$. Suppose $c(0)=0$. Then $c(x+0)=c(x)c(0)-s(x)s(0)=0$. Similarly, $s(x)=0$, which contradicts $a^2+b^2\ne0$. Therefore $c(0)=1$ and $s(0)=0$.

Now \begin{align} c'(x) &=\lim_{h\to0}\frac{c(x+h)-c(x)}{h} \\[6px] &=\lim_{h\to0}\frac{c(x)c(h)-s(x)s(h)-c(x)}{h} \\[6px] &=\lim_{h\to0}\left(c(x)\frac{c(h)-c(0)}{h}-s(x)\frac{s(h)-s(0)}{h}\right) \end{align} Can you go on?

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    Yes, I can - thank you very much! Now, I understand it. – 2017-02-02