what is sine of a real number
The sine function doesn't actually operate on angles, it's a function from the real numbers to the interval [-1, 1] (or from the complex numbers to the complex numbers).
However, it just so happens that it's a very useful function when the input you give it relates to angles. In particular, if you express an angle as a number in radians (in other words, on a scale where an angle of $2\pi$ corresponds to a full circle), it gives you a value that relates to the ratio of two sides of a right-angled triangle that has that angle in one corner.
If that explanation doesn't satisfy you, then you can look at it another way - if you take it that the sine function does take an angle as input and outputs a number, then the differentiability of it relates to how its output changes as you change the angle slightly. If you go far enough in calculus, you'll learn about functions whose inputs and outputs are bizarre multi-dimensional concepts, and as long as the space of bizarre multi-dimensional concepts has the right properties, you can calculate derivatives in a meaningful sense, and if you can get your head around that then differentiating a function of an angle is small fry.
Imagine the unit circle in the usual Cartesian plane: the set of pairs $(x, y)$ where $x$ and $y$ are real numbers. The unit circle is the set of all such pairs a distance of exactly $1$ from the origin.
Imagine a point moving around the circle. As it travels around the circle, it makes an angle of $t$ radians (not degrees!) with the positive $x$-axis. From now on we call the $x$ coordinate the cosine of $t$; and the $y$ coordinate the sine of $t$.
It's as simple as that. If you only remember this one fact you can figure everything else out: the definitions of the trig functions in terms of triangles, the shape of the graphs of the functions, and everything else.
To repeat: $\cos(t)$ and $\sin(t)$ are the $x$ and $y$ coordinates, respectively, of a point on the unit circle that makes an angle of $t$ radians with the origin and the positive $x$-axis.
There's a picture here ... https://en.wikipedia.org/wiki/Unit_circle
How do we get derivative of $\sin(x)$?
- Define $\sin(x)$ to be the series $A(x) = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$
- Define $\cos(x)$ to be the series $B(x) = 1-\frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$
- Some analysis going on. First, you have to convince yourself that these two series do converge. Second, you have to convince yourself that you can differentiate these two series. This part is actually hard.
- Finally, you just differentiate it, and you will see that $\frac{d}{dx}\sin(x) = \cos(x)$.
Notice that there is no cyclic reasoning here. The only trouble is that you have to convince yourself that defining $\sin$ and $\cos$ in this way actually makes sense.
In case you don't like this proof, there is another more elementary proof which can also be found on ProofWiki. It only uses some facts about limits.