f(t) = (a+A*sin(wt + c))/(b+A*cos(wt + c))
d/dt atan(f(t)) = w
(a,b,c,w,A are constants)
t = time
______________________________
Numeric Simulation can show validty of the shown equality.
The way I went so far:
if d/dt atan(f(t)) = w is true,
atan(f(t)) = wt + c , (where c is a constant)
but the problem now is to show, that
f(t) can be expressed in the form "wt + c"
Any help will be appreciated
sin(x)=tan(x)*cos(x) -> sin(x)^2 = tan(x)^2 - tan(x)^2*sin(x)^2 -> (1+tan(x)^2)*sin(x)^2=tan(x)^2 ->sin(x) = tan(x)/sqrt(1+tan(x)^2)
thus
sin(wt+c) = f(t)/sqrt(1+f(t))
similarly
cos(x)=sin(x)/tan(x) -> cos(x)^2=(1-cos(x)^2)/tan(x)^2 -> (tan(x)^2+1)*cos(x)^2=1 -> cos(x)=1/sqrt(1+tan(x)^2)
thus
cos(wt+c) = 1/sqrt(1+f(t))
substituting in your first equation we have:
f(t) = (a*sqrt(1+f(t))+A*f(t))/(b*sqrt(1+f(t))+B) -> b*f(t)*sqrt(1+f(t)) + B*f(t) = a*sqrt(1+f(t))+A*f(t) -> (b*f(t)-a)*sqrt(1+f(t)) + (B-A)*f(t) = 0
Unless I did some mistake it doesn't look possible to me, unless you have some trivial values for the constants of the function f(t).
How did you demonstrate the equality?
Thank you all.
I reached to the same results as Alessandro M. and Dennis F.
only for a = b = 0 it can be shown.
And it turns out, that my simulation runs exactly in this special case.
so problem sollved.
Thanks (and sorry for bothering you )
Setting A=0 and w=1 one sees that the formula is not valid.
- Badges
- Science topic
- Similar topics
- Mathematical Sciences
- Algebra