Invutil said:
I really appreciate your help. Here is what I was looking for:
Find: asin(o) = Φ
Given: o
I have no idea what you're trying to do here, and the picture doesn't help any.
From the two lines above, you are apparently trying to find ##\Phi## and are given o. (BTW, it's a bad idea to use a variable named o or O, because both look like the number 0.)
Invutil said:
?
What's the purpose of bringing in another variable?
Invutil said:
xf = √(1 - yf2)
0≤Φ≤π/2
x0 = 1
y0 = 0
As I said, the picture doesn't help much. For example, what is T
A? It looks like it's the center of the circle, which should be at the origin. From your earlier comment, this is the unit circle, so the hypotenuses of both of your right triangles are 1 unit in length. Neither of the two right triangles has its angle at the origin identified, which makes things much more complicated than they need to be.
For the triangle with the solid lines, with the one vertex at ##(x_2, y_2)##, if its angle at the left vertex is ##\alpha##, then ##x_2 = \cos(\alpha)## and ##y_2 = \sin(\alpha)##. For the triangle with dotted lines, assuming its angle at the left vertex is ##\beta##, then ##x_0 = \cos(\beta)## and ##y_0 = \sin(\beta)##, NOT 1 and 0, respectively, that you show above.
Invutil said:
Then:
lim n→∞ ( Σi=1n ( √( (xi - xi-1)2 + (yi - yi-1)2 ) ) )
= what?
And what are you trying to do?
Invutil said:
y(i) = y
0 + y
fi/n
x(i) = √( 1 - y(i)
2 )
View attachment 113685