=cos24-cos(24+60)-cos72

= cos 24-(cos 24 cos 60-sin 24 sin 60)-cos 72

=cos24-0.5cos24+ (root 3/2)*sin24-cos72

=0.5cos24+ (root 3/2)*sin24-cos72

=(cos24sin30+sin24cos30)-cos72

=sin54-cos72

=cos36-cos72

=cos36-(2(cos36)^2- 1)

=cos36-2(cos36)^2+ 1

= cos 36 *( 1-2 cos 36)+ 1 & lt； - 、

|

Establish an isosceles triangle ABC with A as the fixed point and ∠BAC=36 |

Then it is obvious that the base angles b and c are both 72 |

The bisector drawn from B intersects AC at point D, as shown in the following figure: |

A |

/ \ |

/ \ |

/ D |

/ \ |

B - C |

Abd ABD, BDC is isosceles triangle |

Now let AD= 1 |

Then |

AD=BD=BC= 1 |

AB=2cos36 |

DC=AB-AD=2cos36- 1 |

In triangle BDC, sine theorem |

DC BD |

- = - |

sin∠CBD sin∠C |

Bring the value in and get |

(2 cos 36- 1)/sin 36 = 1/sin 72 |

= & gt(2 cos 36- 1)* sin 72/sin 36 = 1 |

= & gt(2 cos 36- 1)* 2 cos 36 = 1 |

= & gt(2 cos 36- 1)* cos 36 = 0.5->； -& gt； /

Bring in the original formula

cos24 -sin6 -cos72 =0.5

Attachment: (Supplementary)

The above geometric algebra problem only applies to this problem.

Because the value of cos36 (or cos72) is not calculated in the solution;

If there is such a requirement, the following methods can be adopted: (Do not write the specific process)

First, calculate sin 18:

sin(2* 18)=cos(3* 18)

According to the triple angle formula

= & gt

2sin 18*cos 18=4(cos 18)^3-3cos 18

= & gt

4(sin 18)^2+2sin 18- 1=0

= & gt

Sin 18=[ (root number 5)- 1]/4.

= & gtSin36=[ root number (10-2 root number 5)]/4