如題,
求証:sin2x/ [(sinx cosx-1)(sinx 1-cosx)] =cot(x/2) 網友廻答:
- sin2x/ [(sinx cosx-1)(sinx 1-cosx)]=sin2x/ [(sinx-(1-cosx)(sinx 1-cosx)]=sin2x/ [(sin^2x-(1-cosx)^2]=sin2x/ [sin^2x-1-cos^2x 2cosx]=sin2x/ [sin^2x-(sin^2x cos^2x)-cos^2x 2cosx]=sin2x/ [-2cos^2x 2cosx]...
0條評論