Rewrite cos(arcsin(v)) as an algebraic expression in v.
i know that the answer is squroot of 1-v^2
but i don't know the process of how to come to this answer. Thanks for the help!
Hehehe!!
My math skills stop at elementary math. I have no idea how to help you with this problem!
arcsin(v) or arcsin(v/1) is the angle x so that
sin x = v/1
sketch a rightangled triangle where the hypotenuse is 1 and the side opposite angle x is v
let the adjacent side by k
then by Pythagoras,
k^2 + v^2 = 1^2
k^2 = 1 - v^2
k = √(1 - v^2)
so cosx
= cos(sinarc(v)) = k/1 = √(1 - v^2)
To rewrite cos(arcsin(v)) as an algebraic expression in v, we can use the Pythagorean identity and some properties of trigonometric functions.
1. We start by letting θ = arcsin(v). This means that sin(θ) = v.
2. Using the Pythagorean identity (sin^2(θ) + cos^2(θ) = 1), we can substitute sin(θ) with v and solve for cos(θ).
sin^2(θ) + cos^2(θ) = 1
v^2 + cos^2(θ) = 1
cos^2(θ) = 1 - v^2
cos(θ) = ±√(1 - v^2)
3. Since we are working with the arcsin function, θ lies in the interval [-π/2, π/2]. Therefore, cos(θ) is positive in this interval, so we take the positive square root.
cos(θ) = √(1 - v^2)
So, cos(arcsin(v)) can be rewritten as √(1 - v^2).