已知sin(x π 6)=3 4,则sin(5 6π-x)

sin(PI/6+2x)=cos(PI/2-PI/6-2x)=cos(PI/3-2x)=cos(2*(PI/6-x))=1-2*sin(PI/6-x)^2=1-2*(1/4)^2=7/8tan70*c

f(x)=(√3/2)sin2x+(1/2)cos2x+(√3/2)sin2x-(1/2)cos2x+cos2x+1=√3sin2x+cos2x+1=2sin(2x+π/6)+1周期T=2π/2=π对

cos(xπ/3)=cos(π/2-(-xπ/6))=sin(π/6-x)=3/5

由sin(x+π/6)=0.25得cos（x-π/3）=-0.25,原试可整理为cos（x-π/3）+1-cos(x-π/3)*cos(x-π/3)=11/16

-sin(π+x)=sinx-sin(-π+x)=sinx前面项是-1/4由sinx=-sin(-x)得:sin(5π/6-x)=-sin(x-5π/6)=-sin(x+π/6-π)=sin(x+π/

(1)f(x)＝sin(2x＋π/6)＋sin(2x－π/6)＋2cos²x＝sin2xcosπ/6＋cos2xsinπ/6＋sin2xcosπ/6－cos2xsinπ/6＋2cos&sup

1.f(x)=sin(2x+6/π)+2sin^2x=(√3/2)sin2x+(1/2)cos2x+1-cos2x=(√3/2)sin2x-(1/2)cos2x+1=sin(2x-π/6）+1所以f(

多一个sin(2x+π/6)吧!f(x)=sin(2x+π/6)+2cos²x=sin2xcosπ/6+cos2xsinπ/6+1+cos2x=√3/2sin2x+3/2cos2x+1=√3

你的题目是这样的吗已知函数f(X)=2sin(ax-π/6)sin(ax+π/3)(其中a为正常数,x∈R)的最小正周期为π（1）求a的值（2）在△ABC中,若A＜B,且f(A)＝f(B)＝1/2,求

π2x+π/6属于[π/2+2kπ,3π/2+2kπ]时为减区间,所以x属于[π/6+kπ,2π/3+kπ],k属于Z列表：三行2x+π/60π/2π3π/22πx(根据上面一行的值求出x对应的值)f

f(x)=sinxcosπ/6+cosxsinπ/6+sinxcosπ/6-cosxsinπ/6+cosx=sinx+cosx=√2sin(x+π/4)对称轴则sin(x+π/4)=±1x+π/4=k

sin(x+π/6)=1/3sin(5π/6-x)=sin[π-(x+π/6)]=1/3sin^2(π/3-x)=sin^2[π/2-(x+π/6)]=cos^2(x+π/6)=1-sin^2(x+π

∵sin（π/2－x）＝－2sin（π＋x）,∴cosx＝2sinx,∴（cosx）^2＝4（sinx）^2,∴1－（sinx）^2＝4（sinx）^2,∴（sinx）^2＝1/5.∴sinxcosx

f(x)=sin(2x-π/6)=cos[π/2-(2x-π/6)]=cos(2π/3-2x)=cos(2x-2π/3),故f(x)为偶函数,且其关于x=π/3对称,周期为π,因此a的最小整数为π.

∵sin(x+π6)＝14∴sin(x−5π6)+ sin2(π3−x)=sin[π−(x+π6)]+sin2[π2−( π6+x)]=sin(x+π6)+cos2(x+π6)=1

f(x)=2cos(x-π/6)sin(x+π/6)-√3*(sin(x-π/6))^2+sin(x-π/6)cos(x-π/6）=sin2x+sin(π/3)-(√3/2）[1-cos(2x-π/3

cos(5π/6-X)=-cos[π-(5π/6-X)]=-cos(x+π/6)=-sin[π/2-(x+π/6)]=-sin(π/3-X)

(x+π/6)+(5π/6-x)=π,(x+π/6)+(π/3-x)=π/2根据诱导公式：sin(5π/6-x)=sin[π-(x+π/6)]=sin(x+π/6)=1/4sin^2(π/3-x)=s

【参考答案】D[sin(π/2-x)+sin(π-x)]/[cos(-x)+sin(2π-x)]=2009根据诱导公式,化简(cosx+sinx)/(cosx-sinx)=2009左边分子分母同时除以

任何正弦函数,只要系数是1,值域就是[-1,1]