I presume that sin-1x is being used to represent the inverse sin function (I prefer arcsin x to avoid possible confusion). Make use of the trignometirc relationships: cos2θ + sin2θ = 1 ⇒ cosθ =...
Let y = sin(cos-1(2/5))Suppose x = cos-1(2/5): that is, cos(x) = 2/5then sin2(x) = 1 - cos2(x) = 1 - 4/25 = 21/25so that sin(x) = sqrt(21)/5which gives x = sin-1[sqrt(21)/5]Then y = sin(cos-1(2/5)) =...
The inverse sin function I write as arcsin x. Make use of the trignometric relationships: cos2θ + sin2θ = 1 ⇒ cosθ = √(1 - sin2θ) cotθ = cosθ/sinθ = √(1 - (sinθ)2)/sinθ sin(arcsin x) =...
If I read that correctly, you have: cot(sin-1(2/3)) which I understand to mean cot(arcsin(2/3)) which has the value 1/2 x √5 sin(arcsin(x)) = x cos2θ + sin2θ = 1 ⇒ cosθ = √(1 - sin2θ) cotθ...