sin(pi/2)=infinity/infinity


So i have a problem here. I will probably answer myself to this question after some time, but anyways i will post it here. 

So folow the logic i have here.

In picture you can see right-angled triangle. Let angle between c1 and h be α. Then c2/h = sin(α) . If α increases then so do length of c2 and length of h. If α=90 degrees then c2 is with infinity length, and so is h=infinity. 

so we get that sin(pi/2) = infinity/infinity, which should be undefined, but it is 1. Why?




Comments

Anonymous said…
the c2/h definition of sin in flawed, hence the ridiculous result. The better definition of sin is to consider a circle of radius 1 around the origin of a Cartesian plane. The point of the circle at an angle a from the origin is at (cos(a), sin(a)).
7 Jan 2009, 01:55:00
Jēkabs Kārkliņš said…
Thanks Anonymous for reply.
The definition of sin() you provide is flawless, indeed.

I assume that c2/h = sin(a), can be used only on right angled triangle. So, if a>=90, than there is no right angled triangle, and that's why c2/h=sin(a) fails.
14 Jan 2009, 01:34:00
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