Let $f,g:\Bbb R\to \Bbb R$ be functions satisfying
- $\,f(x-y)=f(x)g(y)-f(y)g(x)$
- $g(x-y)=g(x)g(y)+f(x)f(y)$
for all $x,y\in\Bbb R$, and also
- $\lim_{t\to 0} \dfrac{f(t)}{t}=1$.
Show that $f(x)$ is continuous on $\Bbb R$.
These conditions are clearly satisfied by $$\big(g(x),f(x)\big)=(\cos x,\sin x),$$ but I don't know how to prove it.
I converted my comment into a full blown answer.
Putting $x=y$ in first equation we get $f(0)=0$. Putting $x=y=0$ in second equation we get $g(0)=g(0)g(0)$ so that either $g(0)=0$ or $g(0)=1$. If $g(0)=0$ then putting $y=0$ in the first equation in question gives us $f(x) =0$ for all $x$ and this is forbidden in question. Hence $g(0)=1$.
Next interchanging the roles of $x, y$ you can easily show that $f$ is odd, $g$ is even. Based on this we get the usual addition formulas $$f(x+y) =f(x) g(y) +f(y) g(x), g(x+y) =g(x) g(y) - f(x) f(y) \tag{1}$$ The limit equation in question shows that $\lim_{x\to 0}f(x)=0$ so that $f$ is continuous at $0$. Next putting $x=y$ in second equation in question we get $$f^2(x)+g^2(x)=1\tag{2}$$ and combined with $(1)$ this gives the relation $$g(2x)=1-2f^{2}(x)\tag{3}$$ Letting $x\to 0$ we get $\lim_{x\to 0}g(2x)=1$ or $\lim_{x\to 0}g(x)=1$ so that $g$ is also continuous at $0$. Now from equation $(1)$ we can see by letting $y\to 0$ that both $f, g$ are continuous everywhere.
You can go a bit further. The limit in question combined with the equations $(1),(3)$ above shows that $f'(x) =g(x), g'(x) =-f(x) $ so that both $f, g$ satisfy the differential equation $y''+y=0$ and by initial values we get that $f(x) =\sin x, g(x) =\cos x$. Thus the conditions given in the question characterize the circular functions.