1.[根号3(tan12°-根号3)]/[2sin12°(2cos平方12°-1)]

题目

1.[根号3(tan12°-根号3)]/[2sin12°(2cos平方12°-1)]
2.[根号(sin平方2x+cos平方x)]+sin2x+(根号3)*cos2x

答案

1.
[根号3(tan12°-根号3)]/[2sin12°(2cos平方12°-1)]
=(根号3tan12-3)/[2sin12cos24]
=[cos12(根号3tan12-3)]/[2sin12cos12cos24]
=(根号3sin12-3cos12)/[sin24cos24]
=(根号3sin12-3cos12)/[1/2sin48]
=(根号3sin12-3cos12)/[1/2sin(60-12)]
=(根号3sin12-3cos12)/[根号3/4cos12-1/4sin12]
=-(根号3)/4
2.
[根号(sin平方2x+cos平方x)]+sin2x+(根号3)*cos2x
=根号[(sin2x)^2+cos^2(x)]+sin2x+根号3cos2x
=根号[4sin^2(x)cos^2(x)+cos^2(x)]+(sin2x+根号3cos2x)
=根号{cos^2(x)[4sin^2(x)+1]}+2sin(2x+pi/3)
=|cosx|根号[4sin^2(x)+1]+2sin(2x+pi/3)