Why does Sin(a)≈a for small a in radians?

 Introduction

When we were introduced to calculus during 'A' levels (actually 'AO' levels), an investigation into differentiation was used to show that sin(a)≈a for small angles, if the angles were in radians.

At the time, the observation seemed fairly magical: that somehow by picking radians as the units for degrees made it work out. To me, radians always seemed an arbitrary unit, because it wasn't a whole number, but 2π for a complete circle. And the method didn't help to enlighten us, because it involved repeated calculations for sin(a+d)-sin(a) for a⟶0.

Actually, though the reason is very simple. Let's consider a full circle with a unit radius:

 

The distance around the circle (the circumference) is 2π and any part of that distance a around the circle is its angle in radians (since 2π = 360º). Now let's look at a right-angled triangle embedded in a small part of the arc:

The height of the triangle is h, is nearly the same as the length a of the arc around the circle for angle x. And the height of the triangle is the actual meaning of sin(a), since the length a of the arc is the same as the angle in radians, because the radius=1. If we just enlarge that part, we can see that it's very close, but not quite the same.

Therefore, to understand the equivalence doesn't require any advanced notion of calculus, it can be seen directly. Of course, understanding that ∂h=∂a i.e. when a is infintessimal is calculus.