Let x = sin 2 tan - 1 α and y = sin 1 2 tan - 1 4 3 . If S = α ∈ ℝ : y 2 = 1 - x , then ∑ α ∈ S 16 α 3 is equal to _______.

Step-by-step solution

Given x = sin 2 tan - 1 α = sin tan - 1 2 α 1 - α 2 = sin sin - 1 2 α 1 + α 2 = 2 α 1 + α 2         ⋯ i and y = sin 1 2 tan - 1 4 3 = sin tan - 1 1 2 = sin sin - 1 1 5 = 1 5       . . . ii Now, y 2 = 1 - x ⇒ 1 5 = 1 - 2 α 1 + α 2 (from i   &   ii ) ⇒ 1 + α 2 = 5 + 5 α 2 - 10 α ⇒   2 α 2 - 5 α + 2 = 0 ∴   α = 2 , 1 2 ∴   ∑ α ∈