Firstly we need to explain how we measure the angle. The degrees are not helpful any more so we will use radians (I hope it will become clearer later why we choose to do so).

Draw a circle of unit radius and look at O $=(0,0)$, A $=(1,0)$ and B $=(x,y)$ where B is on the circumference of the circle on the first quadrant. Convince yourself that the angle AOB is the length of the arc of the circle from A to B (that's where radians come in).

We aim to formalize this:

Define $\gamma :[-1,1]:\to R^2$ by $f(t)=(\sqrt{1-t^2},t)$ [We just take the positive square root]

Let O $=(0,0)$, A $=\gamma (0)$, B $=\gamma (t)$ where $t\ge 0$ and define the angle AOB to be the length of $\gamma$ from 0 to $t$. Write $\theta (t)$ for this. [Convince yourself that is what we want the angle to be]

Show that $\theta (t)={\int }_0^{t}1/\sqrt{1-x^2}\mathrm{dx}$.

This can also be extended for negative $t$.

Define $\pi$ to be equal to $2\theta (1)$ [Check that it agrees with what what we understand by $\pi$].

Now some more analysis:

Show that $\theta :[-1,1]\to [-\pi /2,\pi /2]$ is strictly increasing. So ${\theta }^{-1}$ exists. Call it sin! [Note that for $0<y<1$ if $\gamma (y)=(x,y)$, then $\sin (\theta (y))=y$, which is what we want it to be! (Draw a picture)]

Extend its domain to the whole of $R$ so that it is periodic.

Show that sin is continuous twice differentiable with $\sin \text{'}\text{'}(x)+\sin (x)=0$ for all real $x$.

Solve this differential equation (using power series) to recover your lecturer's definition of sin.