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why sin(x) is different to sqrt(1-cos(x).^2)

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Luis Isaac
Luis Isaac 2015 年 12 月 30 日
回答済み: Walter Roberson 2015 年 12 月 30 日
Dear, I'm new at matlab programming
I do not undestand why the following results s and s2 are different:
teta=0:pi/10:2*pi;
s=sin(teta);
s2=sqrt(1-cos(teta).^2);
sum(abs(s-s2))
The answer is
ans =
12.6275
But I think that the result should be 0 because sin(x)=sqrt(1-cos(x)^2) Why am I wrong?, where is my mistake?
Sorry for this question and thanks in advance
  1 件のコメント
Luis Isaac
Luis Isaac 2015 年 12 月 30 日
Sorry for the question: The answer is the absolute value: sqrt(x^2)~=x and sqrt(x^2)==abs(x)!!!

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Walter Roberson
Walter Roberson 2015 年 12 月 30 日
sum(sin(teta).^2 + cos(teta).^2 -1) is 0 to within roundoff error.
What you are missing is the adjustment for sign, because sqrt() returns the positive square root only

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