化简½sin²x[(1/tan二分之x)-tan二分之x]+二分之根号三倍cos2x
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解决时间 2021-03-08 16:48
- 提问者网友:沉默的哀伤
- 2021-03-07 16:51
化简½sin²x[(1/tan二分之x)-tan二分之x]+二分之根号三倍cos2x
最佳答案
- 五星知识达人网友:等灯
- 2021-03-07 16:58
解:1/2*sin²x*[1/tan(x/2)-tan(x/2)]+√3/2*cos2x
=1/2*sin²x*[cos(x/2)/sin(x/2)-sin(x/2)/cos(x/2)]+√3/2*cos2x
=1/2*sin²x*[cos²(x/2)-sin²(x/2)]/[sin(x/2)cos(x/2)]+√3/2*cos2x
=sin²x*cosx/sinx+√3/2*cos2x
=1/2*sin2x+√3/2*cos2x
=sin2xcos(π/3)+cos2xsin(π/3)
=sin(2x+π/3)
=1/2*sin²x*[cos(x/2)/sin(x/2)-sin(x/2)/cos(x/2)]+√3/2*cos2x
=1/2*sin²x*[cos²(x/2)-sin²(x/2)]/[sin(x/2)cos(x/2)]+√3/2*cos2x
=sin²x*cosx/sinx+√3/2*cos2x
=1/2*sin2x+√3/2*cos2x
=sin2xcos(π/3)+cos2xsin(π/3)
=sin(2x+π/3)
全部回答
- 1楼网友:由着我着迷
- 2021-03-07 18:33
变形的sinx/2=2cosx/2的tanx/2=2,,,tanx=tan(x/2+x/2)=(tanx/2+tanx/2)/1+tanx/2*tanx/2=4/52,化简的;cosx+sinx/sinx分子分母同处cosx的1+tanx/tanx解的=9/4
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