How does the MATLAB calculate the arctan?
14 views (last 30 days)
Show older comments
Hello all,
I have solved an initial value problem and I have gotten the following equation for that:
theta=(c_0/c_1)- (2/c_1)*atan( exp(-a*c_1*t)*tan((c_0-c_1*theta_0)/2) )
where
c_0=7*pi/6; c_1=0.3; a=0.055; theta_0=0;
and
t=[0:0.01:100];
I do expect the MATLAB returns theta=0 for t=0. In other words what I expect to see is:
theta(1)=0
because for t=0, the first equation can be simplified and as a result we have: theta=theta_0 : independent of c_0,c_1(~=0),and a.
but MATLAB returns something else:
theta(1)=20.9440
I would be grateful if somebody could explain me how I can get what I expect to get?
thanks a lot, Vahid
0 Comments
Accepted Answer
Matt Tearle
on 9 Feb 2012
All inverse trigonometry functions return to a specific limited range, because trig functions are periodic. Hence, if x = 9*pi/2, then sin(x) will be 1, so asin(sin(x)) will be pi/2, not 9*pi/2. That's what's happening here -- atan returns values between -pi/2 and pi/2 (see doc atan):
(c_0-c_1*theta_0)/2 % ans = 1.8326 > pi/2
tan((c_0-c_1*theta_0)/2)
atan(tan((c_0-c_1*theta_0)/2))
atan(tan((c_0-c_1*theta_0)/2)) + pi
0 Comments
More Answers (1)
Wayne King
on 9 Feb 2012
Why do you think it simplifies like that?
for t=0 and theta_0= 0, your expression evaluates to
(c_0/c_1)- (2/c_1)*atan(tan(c_0/2))
which is 20.9440
3 Comments
See Also
Categories
Find more on Uncertainty Analysis in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!