Finding \(\displaystyle{\left({\frac{{{1}}}{{{2}}}}+{\cos{{\frac{{\pi}}{{{20}}}}}}\right)}{\left({\frac{{{1}}}{{{2}}}}+{\cos{{\frac{{{3}\pi}}{{{20}}}}}}\right)}{\left({\frac{{{1}}}{{{2}}}}+{\cos{{\frac{{{9}\pi}}{{{20}}}}}}\right)}{\left({\frac{{{1}}}{{{2}}}}+{\cos{{\frac{{{27}\pi}}{{{20}}}}}}\right)}\)

Marshall Wolf

Marshall Wolf

Answered question

2022-04-06

Finding (12+cosπ20)(12+cos3π20)(12+cos9π20)(12+cos27π20)

Answer & Explanation

fallendreamsit2p

fallendreamsit2p

Beginner2022-04-07Added 14 answers

Let ξ=exp(2πi40). The given product equals
116(1+ξ+ξ1)(1+ξ3+ξ3)(1+ξ9+ξ9)(1+ξ27+ξ27)
or
116ξξ3ξ9ξ27·ξ3-1ξ-1·ξ9-1ξ3-1·ξ27-1ξ9-1·ξ81-1ξ27-1
or (by the telescoping nature and the ξ81=ξ
116ξ1+3+9+27=116ξ40=116

embemiaEffoset4rs

embemiaEffoset4rs

Beginner2022-04-08Added 14 answers

It can be concluded in the following manner:
(12+cosπ20)(12+sinπ20)(12+cos3π20)(12sin3π20)=
=(12+cos9)(12+sin9)(12+cos27)(12sin27)=
=(14+12(cos9+sin9)+12sin18)(14+12(cos27sin27)12sin54)=
=(14+12(sin81+sin9)+12sin18)(14+12(sin63sin27)12cos36)=
=(14+sin45cos36+12sin18)(14+sin18cos4512cos36)=
=(14+5+142+518)(14+51425+18)=
=(18+542+142+58)(18+54214258)=
=(18+542)2(142+58)2=
=164+5162+5325645162132=18116=116

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