Yes. (I assume that by "the intuitive definition based on the trigonometric circle" you mean that (cos t, sin t) is the point at angle t on the unit circle. If

Yes. (I assume that by "the intuitive definition based on the trigonometric circle" you mean that (cos t, sin t) is the point at angle t on the unit circle. If you meant something else, then what follows may not be helpful.) APPROACH 1 If you take the point (1,0) and rotate it anticlockwise through 60º about the origin, you get (cos 60º, sin 60º), by definition. Look at the triangle formed by (0,0), (1,0), and (cos 60º, sin 60º). The angle at the (0,0) vertex is 60º. The other two angles are equal because the triangle is isosceles (rotating a line segment doesn't change its length) so they must also be 60º since the angles in a triangle add up to 180º. So now drop a perpendicular from (cos 60º, sin 60º) down to the horizontal side of the triangle; it lands at (cos 60º, 0), so cos 60º is the distance from (0,0) to that point, and 1-cos 60º is the distance from (1,0) to that point. But those distances are the same, by symmetry, and now we're done. APPROACH 2 Draw the unit circle and inscribe a regular hexagon in it with (1,0) as one vertex. Divide it up into six triangles by drawing radii. As above, these are equilateral triangles. So now look at the horizontal separation between the opposite points (-1,0) and (1,0). Obviously the distance is 2 units. But we can split it up a different way: we have a horizontal distance 1-cos 60º from either of those to the next point around the circle, then 1 from that to the next point, then 1-cos 60º back to the opposite point. So 2 = (1-cos 60º)+1+(1-cos 60º) and again we're done. (But I think any proof of this particular thing is going to be basically about equilateral triangles more than it's about circles. It might be easier first to show that for angles between 0 and 90º the trig functions have their "traditional" definition in terms of right-angled triangles, and then bisect an equilateral triangle.)