I had learnt that:
$\sin \theta = \frac {\text{perpendicular}} {\text{hypotenuse}}$
$\cos \theta = \frac {\text{base}} {\text{hypotenuse}}$
But in unit circles, we find the trigonometric values of obtuse angles. How is that even possible when there's no right angled triangle present?
Answer
A lot of the time, mathematicians will devise something that works for a particular case, and then see if they can extend or generalise it to a bigger area. For example, you can use fairly basic proofs to show that if $n$ and $m$ are integers, then $a^n \times a^m = a^{n + m}$. So mathematicians ask - do they have to be integers? Can we extend the exponentiation rule so that it works for any $n$ and $m$? And if you just take it as given, then it works fine, and then when you devise some alternative ways of expressing exponentiation you can prove it explicitly and it doesn't really break anything.
So, similarly, you can start with $\sin \theta = \frac{\mbox{opposite}}{\mbox{hypotenuse}}$ for $0 \leq \theta \leq \pi$, then you derive a few neat properties of the function, then you draw a circle around your triangle and scale it down so the radius is equal to 1, and then you ask "but what if I used a different angle?" and you can see that you can assign values to the trigonometric functions based on the point on the circle, and then you just confirm that your nice properties still hold - things like $\sin^2 \theta + \cos^2 \theta = 1$ and so forth - and you just say "we will define the sine function to be like this" and it all works nicely.
Which is not to say that it always works nicely - after you make such an extension you still have to prove that it makes sense and doesn't break anything else, otherwise you wind up with things like $1 - 1 + 1 - 1 + \ldots = -\frac{1}{2}$.
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