Want to do is find out how many solutions there are to the following equation in that interval (here

kejpsy

kejpsy

Answered question

2022-01-27

Want to do is find out how many solutions there are to the following equation in that interval (here A is a constant): cos(Acos(sA))=sin(Acos(sA))=Asin(sA),    s[0,2Aπ]
I know it depends on A and I've tested a few cases, but I'm rusty on this trig stuff, so I haven't been able to do it yet

Answer & Explanation

ebbonxah

ebbonxah

Beginner2022-01-28Added 15 answers

For cos(X) to equal sin(X), you must have
X=π4+kπ
So you have
Acos(sA)=π4+kπ
cos(sA)=π4A+kπA
where k is an integer. Since the output of cos is trapped between -1 and 1, for all but finitely many values of k, there are no solutions. There will only be solutions when
1π4A+kπA1
Aπ14kAπ14
For each such k there will be 2 solution values for s in [0,2Aπ) unless for some particular k, π4A+kπA happens to be exactly 1 or -1, and then there will be only one solution value for s.
So count integers between Aπ14 and Aπ14 , and then double. If necessary adjust slightly if π4A+kπA happens to be ±1 for some k.

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