So, this is a trig problem but the catch is you cant use the double angle theorem to solve. the question is, simplify in terms of v: sin(2arcsin[v]) i tried a geometric approach, but it doesnt help much. any help is appreciated!
Let's start with the expression sin(2arcsin[v]). We know that:
sin(2θ) = 2sin(θ)cos(θ)
So, we can rewrite our expression as:
sin(2arcsin[v]) = 2sin(arcsin[v])cos(arcsin[v])
Using the fact that sin(arcsin[x]) = x and cos(arcsin[x]) = sqrt(1 - x^2), we can simplify further:
sin(2arcsin[v]) = 2v*sqrt(1 - v^2)
Therefore, sin(2arcsin[v]) simplifies to 2v*sqrt(1 - v^2) in terms of v.
Using the areas of two triangles works.
Draw a right-angled triangle, vertical leg v, hypotenuse 1; the length of the base will be sqrt(1 - v^2) and the base angle will be arcsin(v).
Add the mirror image of that to form an isosceles triangle with sides 1, 1, 2v and say that the area of the second triangle (1/2)1.1.sin(2arcsin(v)) is double the area of the earlier triangle.