Prove that \(\displaystyle{\frac{{{1}}}{{{{\sin}^{{2}}{\frac{{\pi}}{{{4}{k}+{2}}}}}}}}+{\frac{{{1}}}{{{{\sin}^{{2}}{\frac{{{3}\pi}}{{{4}{k}+{1}}}}}}}}+{\frac{{{1}}}{{{{\sin}^{{2}}{\frac{{{5}\pi}}{{{4}{k}+{2}}}}}}}}+\ldots+{\frac{{{1}}}{{{{\sin}^{{2}}{\frac{{{\left({2}{k}-{1}\right)}\pi}}{{{4}{k}+{2}}}}}}}}={2}{k}{\left({k}+{1}\right)}\)

jermandcryspza3

jermandcryspza3

Answered question

2022-04-02

Prove that
1sin2π4k+2+1sin23π4k+1+1sin25π4k+2++1sin2(2k1)π4k+2=2k(k+1)

Answer & Explanation

sa3b4or9i9

sa3b4or9i9

Beginner2022-04-03Added 14 answers

Using from this answer, we get
k=1n1sin2(2k14b+2)π}=k=1ncsc2(2k12n+1π2)
=k=1nsec2(2n2k+22n+1π2)
=k=1nsec2(k2n+1π)
=n+k=1ntan2(k2n+1π)
=n+n(2n+1)
=2n(n+1)

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