If 0 < x < 1 2 and sin - 1 x α = cos - 1 x β , then a value of sin 2 π α α + β is

Step-by-step solution

Let sin - 1 x α = cos - 1 x β = k i.e. sin - 1 x = k α and cos - 1 x = k β ⇒ sin - 1 x + cos - 1 x = k α + β i.e. k α + β = π 2 ⇒ k = π 2 α + β Now 2 π α α + β = 4 k α = 4 sin - 1 x So sin 2 π α α + β = sin 4 sin - 1 x = 2 sin 2 sin - 1 x cos 2 sin - 1 x = 2 sin sin - 1 2 x 1 - x 2 cos sin - 1 2 x 1 - x 2 = 2 sin sin - 1 2 x 1 - x 2 cos cos - 1 1 - 2 x 2 = 4 x 1 - x 2 1 - 2 x 2