Finding \(\displaystyle\lim_{{{x}\to{0}}}{\frac{{{\sin{{\left(\pi{{\cos}^{{{2}}}{\left({\frac{{{x}}}{{{2}}}}\right)}}\right)}}}}}{{{\sin{{\left({\sin{{\left({x}\right)}}}\right)}}}}}}\) I have the following solution, only

Camila Glenn

Camila Glenn

Answered question

2022-04-07

Finding limx0sin(πcos2(x2))sin(sin(x))
I have the following solution, only the first equality of which bothers me:
limx0sin(πcos2(x2))sin(sin(x))=limx0sin(πsin2(x2))sin(sin(x))=limx0πsin(x2)2cos(x2).2sin(x2)cos(x2)sin(2sin(x2)cos(x2)).sin(πsin2(x2))πsin2(x2)=0
What is the justification for replacing cos2 with sin2?

Answer & Explanation

star04iks7

star04iks7

Beginner2022-04-08Added 14 answers

cos2(x)=1sin2(x), so sin(πcos2(x))=sin(ππsin2(x))=sin(πsin2(x)) ,since sin(πy)=sin(y)

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