✨ ベストアンサー ✨
z1/z2 = (-1 + i)/(√(3) + i) = (-1 + i)(√(3) - i)/4
= (1 - √(3))/4 + {(1 + √(3))/4}i ...①
z1 = -1 + i = √(2)(cos(3π/4) + isin(3π/4))
z2 = √(3) + i = 2(cos(π/6) + isin(π/6))
よって
z1/z2 = (1/√(2))(cos(3π/4 - π/6) + isin(3π/4 - π/6))
= (1/√(2))(cos(7π/12) + isin(7π/12)) ...②
①,②を比較して
(1/√(2))cos(7π/12) = (1 - √(3))/4
(1/√(2))sin(7π/12)) = (1 + √(3))/4
よって
cos(7π/12) = (√(2) - √(6))/4
sin(7π/12) = (√(2) + √(6))/4

ありがとうございます!もう1つ質問載せたのでそちらもよければお願いします…