The value sin θ used to be defined as the length of the semichord subtending an

Question

angle θ at the centre of a circle of radius 1.
In the diagram to the right, the chord AB of a circle of radius 1 subtends an
angle 2θ at the centre O. We know that the perpendicular bisector of the chord
passes through the centre O. Prove that sin θ = 1/2 AB .

Answer 1

To prove that sin θ = 1/2 AB, we can consider triangle AOB.

Since the perpendicular bisector of the chord AB passes through the center O, we know that triangle AOB is an isosceles triangle with AO = BO = 1.

Let x be the length of the chord AB. Since the angle subtended by the chord at the center is 2θ, we can write:

sin θ = (1/2) x/1
sin θ = (1/2) x

Thus, sin θ = 1/2 AB. This proves that sin θ = 1/2 AB.