高校数学の質問スレ Part411

tan(π/8) = √2 - 1　を使うなら

tan(3/2) = t(3-tt)/(1-3tt) = 1/tan((π-3)/2)

ところで 0<θ<π/8 では
　θ < tanθ < {8(√2 -1)/π}θ = (19/18)θ,
　tan((π-3)/2) < 19(π-3)/36 < 3/40,
　tan(3/2) = t(3-tt)/(1-3tt) > 40/3,
　tan(1/2) = t > 49/90 = 0.5444444

t(1+tt) - (1-tt) = t^3 + t^2 + t - 1
　= 1639/90^3 + (t - 49/90){tt + (139/90)t + 14911/(90^2)}
　> 1639/90^3,

t > (1-tt)/(1+tt) = cos(1),