sin(π/10) = sin (180/10) = sin 18

Let A = 18

=> 5A = 5*18

=> 5A = 90

=> 2A + 3A = 90

=> 2A = 90 - 3A

=> sin 2A = sin(90 - 3A)

=> sin 2A = cos 3A

=> sin 2A - cos 3A = 0

=> 2*sin A*cos A - {4*cos³ A - 3*cos A} = 0

=> 2*sin A*cos A - 4*cos³ A + 3*cos A = 0

=> cos A (2*sin A - 4*cos² A + 3) = 0

=> cos A = 0 ≠ cos 18

Again, 2*sin A - 4*cos² A + 3 = 0

=> 2*sin A - 4(1 - sin² A) + 3 = 0

=> 2sin A - 4 + 4sin² A + 3 = 0

=> 4sin² A + 2sin A - 1 = 0

=> sin A = [-2 ± √{(-2)² - 4*4*(-1)}]/(2*4)

=> sin A = [-2 ± √{4 + 16}]/8

=> sin A = [-2 ± √20]/8

=> sin A = [-2 ± 2√5]/8

=> sin A = [-1 ± √5]/4

since A lies in first quadrant, sin A > 0

So, sin A = [-1 + √5]/4