Question

based on the diagram, which equation can be simplified to derive the cosine sum identity?
√(cos(u + v) - 1)² + (sin(u + v) - 0)² = √(cos(u) - 1)² + (sin(u) - 0)²
√(cos(u + v) - 1)² + (sin(u + v) - 0)² = √(cos(-v) - 1)² + (sin(-v) - 0)²
√(cos(u + v) - 1)² + (sin(u + v) - 0)² = √cos(u)-cos(-v)² + sin(u)-sin(-v)²
√(cos(u + v) - 1)² + (sin(u + v) - 0)² = √(cos(u + v) - cos(u))² + (sin(u + v) - sin(u))²

Explanation:

Step1: Recall the distance formula

The distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). In the unit - circle, to derive the cosine sum identity \(\cos(u + v)=\cos u\cos v-\sin u\sin v\), we use the fact that the length of the arc \(PR\) (where \(P(1,0)\) and \(R(\cos(u + v),\sin(u + v))\)) and the length of the arc \(QS\) (where \(Q(\cos u,\sin u)\) and \(S(\cos(-v),\sin(-v))\)) are equal. Since the distance between two points on the unit circle corresponding to equal - length arcs is the same.
The distance between \(R(\cos(u + v),\sin(u + v))\) and \(P(1,0)\) is \(d_1=\sqrt{(\cos(u + v)-1)^2+(\sin(u + v)-0)^2}\).
The distance between \(Q(\cos u,\sin u)\) and \(S(\cos(-v),\sin(-v))\) is \(d_2=\sqrt{(\cos u-\cos(-v))^2+(\sin u-\sin(-v))^2}\).
Since \(d_1 = d_2\) (because of the symmetry and arc - length equality in the unit circle for the purpose of deriving the cosine sum formula).

Answer:

\(\sqrt{(\cos(u + v)-1)^2+(\sin(u + v)-0)^2}=\sqrt{(\cos u-\cos(-v))^2+(\sin u-\sin(-v))^2}\)