based on the diagram, which equation can be simplified to derive the cosine sum identity?\n√(cos(u + v)…

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

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

Answer

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})