F.4 TRIGONOMETRY M2

2011-03-16 4:00 am
1.prove that cos2A+cos2B-cos2C-1=-4sinAsinBcosC

2.show that sin20sin40sin60sin80= 3/16

回答 (1)

2011-03-16 4:23 am
✔ 最佳答案
1)cos2A + cos2B - cos2C - 1= 2 cos(A+B) cos(A-B) - (2cos²C - 1) - 1= 2 [cos(A+B) cos(A-B) - cos²C]= 2 [- cosC cos(A-B) - cos²C]= - 2 cosC [cos(A-B) + cosC]= - 2 cosC [cos(A-B) - cos(A+B)]= - 2 cosC [cosAcosB + sinAsinB - (cosAcosB - sinAsinB)]= - 2 cosC (2sinAsinB)= - 4 sinA sinB cosC
2)sin20 sin40 sin60 sin80= √3/2 sin20 sin40 sin80= (√3/2) (-1/2)(cos60 - cos20)sin80= (√3/2) [(-1/4)sin80 + (1/2)sin80cos20]= (√3/2) [(-1/4)sin80 + (1/4)(sin100 + sin60)]= (√3/2) [(-1/4)sin80 + (1/4)(sin80 + sin60)]= (√3/2) (1/4)sin60= (√3/2) (1/4)(√3/2) = 3/16


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