搜狗做题网已知cos(六分之π-x)=三分之根号三,求cos(六分之-
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后面的sin应该有平方(1)f(x)=√3sin（2X－π/6）+2sin²(x-π/12)=√3sin（2X－π/6）+1-cos(2x-π/6)=3sin（2X－π/6）-cos(2x-
f(x)=cosx-cos(x+π/2)=cosx+sinx=3/4sin^2x+cos^2x+2sinxcosx=9/162sinxcosx=sin2x=9/16-1=-7/16
1、①增：2kπ－π≤4x－π/6≤2kπ求出x范围；②减2kπ≤4x－π/6≤2kπ＋π求出x范围.2、最大值是2,此时4x－π/6=2kπ,最小是－2,此时4x－π/6=2kπ＋π.求出x的范围,
（2/3+5/6）X=3/5
f(x)=根号3cos^x+sinxcosx-根号3/2=根号3*(1+cos2x)/2+sin2x/2-根号3/2所以f(派/8)=根号3*(1+cos派/4)/2+sin(派/4)/2-根号3/2
再问：能在详细一点吗？再答：你哪步没看懂，基本没跳步呀再问：哦，不是，我看错了，挺详细的，谢谢！再答：不言谢！
很高兴为你解答~5/6-3/8=1/3x11/24=1/3xx=11/8
∵cos（x-π/6）=-√3/3∴cosx+cos（x-π/3）=cosx+cosxcosπ/3+sinxsinπ/3=cosx+（1/2）cosx+（√3/2）sinx=（3/2）cosx+（√3
cos(x-π/6)=-√3/3cosx+cos(x-π/3)=cos(x-π/6+π/6)+cos(x-π/6-π/6)=cos(x-π/6)cosπ/6-sin(x-π/6)sinπ/6+cos(
因为cos(a+b)=cosacosb-sinasinbcos(a-b)=cosacosb+sinasinb相加得cos(a+b)+cos(a-b)=2cosacosb即cosacosb=[cos(a
cos(π/6)=cos30度=(根号3)/2.sin(π/6)=sin30度=1/2.π,π/2,π/3,π/4...之类的,化为角度再求.
∵cos（π/8-α）=√3/3∴π/8-α=±π/6+2kπ,k∈Zα=7π/24+2kπ或-π/24+2kπ,k∈Z当α=7π/24+2kπ时,原式=sin^2（7π/24+2kπ-π/6）-co
f(x)=1-2sin^2(x)+1-sin^2(x)-4sinx+2=-3sin^2(x)-4sinx+4.f(π/6)=5/4.因为-1《sinx《1.当sinx=2/3时,f(x)min=0当s
2tanx乘以sinx=3即2sinx/cosx*sinx=3∴2sin²x=3cosx∴2（1-cos²x)=3cosx∴2cos²x+3cosx-2=0解得:cosx
90-16.778
设函数f（x）=cos（2x派/3）sin^2x?1：求f（x）的值域和最小正故值域为[1/2-根号3除以2,1/2根号3除以2],最小正周其为π
17π/12＜x＜7π/4,得5π/3＜x+π/4＜2πcos(x-π/4)=cos[(x+π/4)-π/2]=sin(x+π/4)=-√[1-sin²(x+π/4)]=-√[1-(3/5)