三角函数的，COS72°-----COS36°

cos72 - cos36
= cos(54+18) - cos(54-18)
= cos54 * cos18 - sin54 * sin18 - [cos54 * cos18 + sin54 * sin18]
= -2sin54*sin18
= -2cos36*sin18*cos18 / cos18
= -cos36sin36 / cos18
= -1/2 * 2cos36sin36 / cos18
= -1/2 * sin72 / cos18
= -1/2 * cos18 / cos18
= -1/2

原式=(sin36cos72-sin36cos36)/sin36
=(sin108-sin36-sin72)/2sin36
=(sin72-sin36-sin72)/2sin36
=-sin36/2sin36
=-1/2

答：COS72°-COS36°
=cos(54°+18°)-cos(54°-18°)
=[cos54°*cos18°-sin54°*sin18°]-[cos54°*cos18°+sin54°*sin18°]
=-2sin54°*sin18°
=-2cos36°*sin18°
=-cos36°*(2sin18°*cos18°)/cos18°
=-cos36°*sin36°/cos18°
=-(2cos36°*sin36°)/(2cos18°)
=-sin72°/(2cos18°)
=-cos18°/(2cos18°)
=-1/2

cos36°cos72°(添加2sin36°)
=2sin36° cos36°cos72°/2sin36°
=sin72°cos72°/2sin36°
=2 sin72°cos72°/2^2 sin36°
=sin144°/2^2sin36° [公式：sin（π－α）＝sinα)]
=sin36°/2^2sin36°
=1/2^2

原题到底是COS72°- COS