Prove that \(\displaystyle{\frac{{{1}}}{{{90}}}}{\sum_{{{n}={1}}}^{{{90}}}}{2}{n}\cdot{\sin{{\left({\left({2}{n}\right)}^{\circ}\right)}}}={\cot{{\left({1}^{\circ}\right)}}}\)

Zoie Phillips

Zoie Phillips

Answered question

2022-03-31

Prove that
190n=1902nsin((2n))=cot(1)

Answer & Explanation

Cason Harmon

Cason Harmon

Beginner2022-04-01Added 8 answers

S=n=1902(91n)sin(2n)=n=1902k=1nsin(2k)
Since:
2(sin1)k=1nsin(2k)
=k=1n(cos((2k1))cos((2k+1)))=cos1cos((2n+1))
it follows that:
S=1sin1(90cos1n=190cos((2n+1)))
n=190cos((2n+1))=2cot1
hence S=92cot1. On the other hand, since:
n=1902sin(2n)=2cot1
always by the same trick, it follows that:
n=1902nsin(2n)=(91292)cot1=90cot1
as wanted.

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