thấy 2π/5 + 3π/5 = π => sin(2π/5) = sin(3π/5) <=> 2sin(π/5).cos(π/5) = 3sin(π/5) - 4sin³(π/5) <=> 2cos(π/5) = 3 - 4sin²(π/5) = 3 - 4 + 4cos²(π/5) <=> 4cos²(π/5) - 2cos(π/5) - 1 = 0 <=> cos(π/5) = (1-√5)/4 (loại vì cosπ/5 > 0) hoặc cos(π/5) = (1+√5)/4 Vậy cos(π/5) = (1 + √5)/4 => cos(2π/5) = 2cos²(π/5) - 1 = 2.(6+2√5)/16 - 1 = (√5-1)/4
thấy
$2π/5 + 3π/5 = π
$$=> sin(2π/5) = sin(3π/5)
$ $<=> 2sin(π/5).cos(π/5) = 3sin(π/5) - 4sin³(π/5)
$$<=> 2cos(π/5) = 3 - 4sin²(π/5) = 3 - 4 + 4cos²(π/5)
$$<=> 4cos²(π/5) - 2cos(π/5) - 1 = 0
$ $<=> cos(π/5) = (1-√5)/4 (loại vì cosπ/5 > 0) hoặc cos(π/5) = (1+√5)/4
$ Vậy
$cos(π/5) = (1 + √5)/4
$$=> cos(2π/5) = 2cos²(π/5) - 1 = 2.(6+2√5)/16 - 1 = (√5-1)/4
$