The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.
You can use the arctangent or the reverse tangent to solve for x, which is denoted by arctan or tan^-1. If tan [x] = 3, then arctan [3] = x. This applies to all trigonometric functions (ex. if sin [x] = 94, then arcsin [94] = x. Punch that into your calculator and the answer will be: arctan [3.0] = 71.565 (degrees) arctan [3.0] = 1.249 (radians)
Arcsin
Yes, it is called arcsin.
sin(arcsin(2/3)) = 2/3, since sin is the inverse function of arcsin.
arcsin(.75)≈0.848062079
arcsin(0.329) = 0.335 radians = 19.21 degrees
Yes, both arcsin and sine inverse are the same.
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) = x Then: cot(arcsin(x)) = √(1 - (sin(arcsin(x))2)/sin(arcsin(x)) = √(1 - x2)/x ⇒ cot(arcsin(4/7)) = √(1 - (4/7)2)/(4/7) = √(49/72 - 16/72) ÷ 4/7 = √(49 - 16) x 1/7 x 7/4 = 1/4 x √33
That will depend on exactly how the equation is formed. In many cases, you can apply the inverse function to the outside first. Here is an example:sin(ln(x)) = ... To solve for "x", FIRST apply the inverse function of the sine (i.e., arcsin) to both sides of the equation. Next, apply the inverse of the natural logarithm to both sides. In this case, the exponential function (raise "e" to the power of the entire expression on both sides).
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 = √(1 - sin2θ) cotθ = cosθ/sinθ = √(1 - sin2θ)/sinθ sin(arcsin x) = x Then: cot(arcsin(x)) = √(1 - sin2(arcsin(x))/sin(arcsin(x)) = √(1 - x2)/x ⇒ cot(arcsin(2/3)) = √(1 - (2/3)2)/(2/3) = √(9/32 - 4/32) ÷ 2/3 = √(9 - 4) x 1/3 x 3/2 = 1/2 x √5
arc sine is the inverse function of the sine function so if y = sin(x) then x = arcsin(y) where y belongs to [-pi/2, pi/2]. It can be calculated using the Taylor series given in the link below.
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 = cosθ ÷ sinθ ⇒ cot(arcsin(2/3)) = cos(arcsin(2/3)) ÷ sin(arcsin(2/3) = √(1 - sin2(arcsin(2/3))) ÷ sin(arcsin(2/3) = √(1 - (2/3)2) ÷ (2/3) = 1/3 x √(9 - 4) x 3/2 = 1/2 x √5 As the reciprocal trignometric functions have separate names, eg 1/tan x = cot x, the use of the -1 "power" to indicate the inverse function is possible. However, to avoid any possible confusion, I prefer to use the arc- prefix to indicate the inverse function.
arcsin(1) arccos(0)