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Question Number 110545 by peter frank last updated on 29/Aug/20

Commented by Dwaipayan Shikari last updated on 29/Aug/20

(1/2)(√(2−(√(2+(√3)))))

1222+3

Commented by Dwaipayan Shikari last updated on 29/Aug/20

((e^((πi)/(24)) −e^((−πi)/(24)) )/(2i))

eπi24eπi242i

Answered by Dwaipayan Shikari last updated on 29/Aug/20

1+cos(π/6)=2cos^2 (π/(12))  (1+((√3)/2))=2cos^2 (π/(12))  (√((1/2)(1+((√3)/2))))=cos(π/(12))=(1/2)(√(2+(√3)))  1−cos(π/(12))=2sin^2 (π/(24))  1−(1/2)(√(2+(√3)))=2sin^2 (π/(24))  (√((1/2)(1−(1/2)(√(2+(√3))))))=sin(π/(24))  (1/2)(√(2−(√(2+(√3)))))=sin(π/(24))

1+cosπ6=2cos2π12(1+32)=2cos2π1212(1+32)=cosπ12=122+31cosπ12=2sin2π241122+3=2sin2π2412(1122+3)=sinπ241222+3=sinπ24

Commented by peter frank last updated on 29/Aug/20

thank you

thankyou

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