Question

函数y=根号3sinx*cosx+3(cosx)^2-3/2

Answer 1

y=√3sinxcosx+3cos^2x-3/2
=(√3/2)sin2x+(3/2)(2cos^2x-1)]
=(√3/2)sin2x+(3/2)cos2x
=√3[(1/2)sin2x+(√3/2)cos2x
=√3(sin2xcosπ/3+cos2xsinπ/3)
=√3sin(2x+π/3)
不知道求什么,只能给你做到这里了,

Answer 2

http://zhidao.baidu.com/question/135614378.html?fr=qrl&cid=983&index=1&fr2=query
我以前做过 请参考

Answer 3

y=√3sinx*cosx+3(cosx)^2-3/2
=（√3/2)（2sinxcosx)+3*(1+cos2x)/2-3/2
=(√3/2)sin2x+3cos2x/2
=√3(sin2x/2+√3cos2x/2)
=√3(sin2xcos60°+cos2xsin60°）
=√3sin(2x+60°）。

Answer 4

y=√3sinx*cosx+3(cosx)^2-3/2
=√3/2 ×2sinx*cosx+3×（cos2x+1)/2 -3/2
=√3/2 sin2x+3/2 cos2x +3/2-3/2
=√3/2 sin2x+3/2 cos2x
=√3 [ 1/2 sin 2x + √3/2 cos 2x)
=√3[cos 60°sin 2x +sin 60°cos 2x]
=√3 sin (2x+60°)
希望你明白！