✔ 最佳答案
1)sin²x + sin2x = m + 2cos²x(1 - cos2x)/2 + sin2x = m + (cos2x + 1)1 - cos2x + 2sin2x = 2m + 2cos2x - 22m = 2sin2x - 3cos2x + 3m = sin2x - (3/2)cos2x + 3/2
2)- cos(A+B) cos(A -B) + sin²C= - cos(A+B) cos(A -B) + 1 - cos²C= - cos(A+B) cos(A -B) + 1 - cos²(A+B)= 1 - cos(A+B) ( cos(A -B) + cos(A+B) )= 1 - cos(A+B) ( cosAcosB + sinAsinB + cosAcosB - sinAsinB ) = 1 - cos(A+B) (2cosAcosB)= 1 + 2 cosA cosB cosC
2011-10-16 20:11:40 補充:
Corrections :
1)
sin²x + sin2x = m + 2cos²x
(1 - cos2x)/2 + sin2x = m + (cos2x + 1)
1 - cos2x + 2sin2x = 2m + 2cos2x + 2
2m = 2sin2x - 3cos2x - 1
m = sin2x - (3/2)cos2x - 1/2
2011-10-16 20:13:15 補充:
- cos(A + B)cos(A - B) + (sinC)^2
= 1 + 2cosAcosBcosC instead of 2 + 2cosAcosBcosC.