F.4 AM

As follows:

圖片參考：http://i238.photobucket.com/albums/ff245/chocolate328154/Maths020.jpg?t=1197112424


2007-12-08 21:14:28 補充：
First of all, cos^2 B = 1 - sin^2 BThen 0<= cos ( A - C ) <= 1 ( sorry that I made a mistake here ) because in the triangle, A and C may equal, then it becomes cos 0 which is 1;

2007-12-08 21:14:42 補充：
but even if they are not equal, it is still > 0 becuase the difference of 2 angles in a triangle cannot exceed 90*.

2007-12-08 21:21:05 補充：
Very sorry, something wrong with the above. cos^2 B = 1 - sin^2 B => fine. But -1 < cos ( A - C ) <= 1 becuase the difference between 2 angles cannot = 180*. Then using the product to sum formula, you can arrive at (1/2)[cos(A-C)-cos(A+C)].

2007-12-08 21:24:17 補充：
Making use of ai), you can find that:2cos^2 B + cosB-1=1-cos(A-C)As mentioned,-1 < cos ( A - C ) <=1, so 1-cos(A-C) must >=0 and hence the prove

2007-12-08 23:21:53 補充：
-cos(A+C)=-cos(180*-B)=-(-cosB)=cosB [ since in the 2nd quad., cos is -VE]

2007-12-08 23:29:07 補充：
This's not an easy Q. I've thought for ages before I could solve it!!!