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How to get the solutions of inverse cosine (acos) in the interval [0, 2π]?

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Giuseppe
Giuseppe 2022 年 3 月 7 日
編集済み: Torsten 2022 年 3 月 7 日
Hi guys!
I've an expression like this one: "cos_omega = ...".
I know that the matlab function acos return only the solution in the interval [0, π].
How can I obtain the two solutions, i.e. the first one in the interval [0, π] and the second one in the interval [π, 2π]?
Is there a matlab function to perform this task?

採用された回答

Torsten
Torsten 2022 年 3 月 7 日
編集済み: Torsten 2022 年 3 月 7 日
Is there a matlab function to perform this task?
No.
But if you want to work within the branch in [0,pi] as well as in [pi,2*pi], you may define your own acos-function as
Acos_User = @(x) [acos(x),2*pi-acos(x)]
where x is given in radians.
I returns a (1x2) vector with both values.

その他の回答 (1 件)

Davide Masiello
Davide Masiello 2022 年 3 月 7 日
編集済み: Davide Masiello 2022 年 3 月 7 日
n = 10;
k = 1:n;
omegas = sort([k*2*pi-acos(cos_omega),k*2*pi+acos(cos_omega)])'
yields solutions up to [(2*n-1)*pi 2*n*pi]. For the case above
omegas =
5.6397
6.9267
11.9229
13.2099
18.2061
19.4931
24.4892
25.7762
30.7724
32.0594
37.0556
38.3426
43.3388
44.6258
49.6220
50.9090
55.9052
57.1922
62.1884
63.4754
>> cos(omegas)
ans =
0.8000
0.8000
0.8000
0.8000
0.8000
0.8000
0.8000
0.8000
0.8000
0.8000
0.8000
0.8000
0.8000
0.8000
0.8000
0.8000
0.8000
0.8000
0.8000
0.8000

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