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Question Number 208252 by efronzo1 last updated on 09/Jun/24

 cos (((2π)/(21)))cos (((4π)/(21)))cos (((8π)/(21)))cos (((10π)/(22)))cos (((16π)/(21)))cos (((20π)/(21)))=?

cos(2π21)cos(4π21)cos(8π21)cos(10π22)cos(16π21)cos(20π21)=?

Answered by som(math1967) last updated on 09/Jun/24

 let (π/(21))=x  cos2xcos4xcos8xcos10xcos16xcos20x  cosxcos2xcos4xcos8xcos10xcos5x  [21x=π⇒cos20x=−cosx,cos16x=−cos5x]  (1/(4sinxsin5x))×sin2xcos2xcos4xcos8xsin10xcos10x  =(1/(64sinxsin5x))×2sin8xcos8x×2sin10xcoss10x  =(1/(64sinxsin5x))×sin16xsin20x  =(1/(64sinxsin5x))×sin5x×sinx  [21x=π ⇒sin16x=sin5x,sin20x=sinx]  =(1/(64))

letπ21=xcos2xcos4xcos8xcos10xcos16xcos20xcosxcos2xcos4xcos8xcos10xcos5x[21x=πcos20x=cosx,cos16x=cos5x]14sinxsin5x×sin2xcos2xcos4xcos8xsin10xcos10x=164sinxsin5x×2sin8xcos8x×2sin10xcoss10x=164sinxsin5x×sin16xsin20x=164sinxsin5x×sin5x×sinx[21x=πsin16x=sin5x,sin20x=sinx]=164

Commented by efronzo1 last updated on 09/Jun/24

Commented by som(math1967) last updated on 09/Jun/24

yes

yes

Commented by Ghisom last updated on 09/Jun/24

I get the same result

Igetthesameresult

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