How does the MATLAB calculate the arctan?

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Vahid
Vahid 2012-2-9
编辑： Marc 2013-10-7
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

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采纳的回答

Matt Tearle
Matt Tearle 2012-2-9
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

更多回答（1 个）

Wayne King
Wayne King 2012-2-9
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
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Vahid
Vahid 2012-2-9
Since we have: atan(tan(arg))=arg and because of that I think the answer should be
(c_0/c_1)- (2/c_1)*(c_0/2)
which is equal to 0!
Wayne King
Wayne King 2012-2-9
Oh, I see :), yes, then what Matt said.

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