三角證明題

RHS
= 1 - 2 cos A cos B cos C
= 1 - [cos(A+B) + cos(A-B)] cos C
= 1 - cos(A+B) cos C - cos(A-B) cos C
= 1 - (1/2)[cos(A+B+C) + cos(A+B-C)] - (1/2)[cos(A-B+C) + cos(A-B-C)]
= 1 - (1/2)cos pi - (1/2)cos(pi - 2C) - (1/2)cos(pi - 2B) - (1/2)cos(2A - pi)
= 1 + 1/2 + (1/2)cos 2C + (1/2)cos 2B + (1/2)cos 2A
= 3/2 + (1/2)(2 cos^2 C - 1) + (1/2)(2 cos^2 B - 1) + (1/2)(2 cos^2 A - 1)
= cos^2 A + cos^2 B + cos^ C
= LHS

2012-10-13 18:54:36 補充：
Sorry, 尾二果行打漏個 2。即
= cos^2 A + cos^2 B + cos^2 C
= LHS

這個有沒有比較短？
cos²A＋cos²B＋cos²C
＝ ( 1＋cos 2A／2 )＋( 1＋cos 2B／2 )＋( 1＋cos 2C／2 )
＝ 3/2＋1/2．( cos 2A＋cos 2B＋cos 2C )
＝ 3/2＋1/2．［2 cos(A＋B) cos(A－B) ＋cos 2C］
＝ 3/2＋1/2．［- 2 cos C cos(A－B) ＋2 cos² C－1］
＝ 3/2＋1/2．［- 2 cos C（cos(A－B) ＋cos(A＋B)）－1］
＝ 3/2＋1/2．［- 2 cos C（2cosAcosB）－1］
＝ 1 － 2cosAcosB cos C

ABC 是一三角形，證明
Cos^2 A+Cos^2 B+Cos^2 C=1－2Cos ACos BCos C
Sol^
2CosACosB
=(CosACosB－SinASinB)+(CosACosB+SinASinB)
=Cos(A+B)+Cos(A－B)
=－CosC+Cos(A－B)
Cos^2 A+Cos^2 B+Cos^2 C+2CosACosBCosC
=Cos^2 A+Cos^2 B+Cos^2 C+CosC[－CosC+Cos(A－B)]
=Cos^2 A+Cos^2 B－Cos(A+B)Cos(A－B)
=Cos^2 A+Cos^2 B－(Cos^2 ACos^2 B－SIN^2 ASIN^2 B)
=Cos^2 A+Cos^2 B－[Cos^2 ACos^2 B－(1－Cos^2 A)(1－Cos^2 B)]
=Cos^2 A+Cos^2 B－[Cos^2 ACos^2 B－(1－Cos^2 A－Cos^2 B
+Cos^2 ACos^2 B)]
=Cos^2 A+Cos^2 B－(Cos^2 ACos^2 B－1+Cos^2 A+Cos^2 B－Cos^2 ACos^2 B)
=1
So
Cos^2 A+Cos^2 B+Cos^2 C=1－2Cos ACos BCos C