已知f(x)=根号3*sin2x 2cos^2 x m(m∈R),
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a=(√3sinx/2,cosx/2),b=(cosx/2,-cosx/2)f(x)=a·b=√3sin(x/2)cos(x/2)-(cosx/2)^2=(√3/2)sinx-(1/2)cosx-1/
y=sin(x/2)+√3cos(x/2)=2sin(x/2+π/3)x=4kπ+π/3(k为整数)时取最大值2
sin2α+根号3倍cos2α=2sin(2α+π/3)=1sin(2α+π/3)=1/22α+π/3=5π/6α=π/4f(x)=2xcos(α-π/4)+sin(α+π/4)=2x+1a1=1a2
f(sin2θ)-f(-sin2θ)=1−sin2θ-1+sin2θ=|sinθ-cosθ|-|sinθ+cosθ|.∵θ∈(5π4,3π2),∴-1<sinθ<-22<cosθ<0.∴cosθ-si
f(sin(x/2))=cosx+1=1-2(sin(x/2))^2+1=2-2(sin(x/2))^2令y=sin(x/2)则f(y)=2-2y^2令y=cos(x/2)f(cos(x/2))=2-
分析:先利用二倍角公式和两角和公式对函数解析式进行化简,得到一个角的一个三角函数的形式,然后再求出函数h(x)的解析式,再根据正弦函数的对称性和t的范围求出t的值.∵y=【2sin²(x+π
f(x)=sin(x/2)+√3cos(x/2)+1=2sin(x/2+π/3)+1T=2π/(1/2)=4π因为正弦函数的值域是[-1,1]所以f(x)∈[-1,3]
1+sin2a=(sina)^2+2sinacosx+(cosa)^2=(sina+cosa)^2cos2a=(cosa)^2-(sina)^2=(sina+cosa)(cosa-sina)所以分子=
f(x)=sin2(x+π)+根号3sin(x+π)sin(π-x)-1\2=sin2x-根号3sin²x-1/2=sin2x+根号3/2cos2x-1=根号7/2sin(2x+γ)-1co
1)aXb=cos(2/3x)cos(2/x)-sin(2/3x)(sin(2/x)=cos(2/3x+2/x)=cos(8/3x)=0所以8/3x∈π/2+kπ,k∈Z即x∈3π/16+3kπ/8,
1、f(sin2)+f(sin(-2))=√(1-sin2)+√[1-sin(-2)]=√(1-sin2)+√(1+sin2)1-sin2=(sin1)^2+(cos1)^2-2sin1cos1=(s
f(x)=sinxsin(xπ/2)=sinxcosx故f(x)的最小正周期是2πf^2(x)=(sinxcosx)^2=1sin2xsin2x=-7/16sin(xπ/2)=cosxf
sin2(x-45°)=-cos2x=2sin^2x-1=2-√5
函数f(x)=(sinx)^3cosx+(cosx)^3sinx+√3(sinx)^2=sinxcosx[(sinx)^2+(cosx)^2]+√3(sinx)^2=sinxcosx+√3(sinx)
定义域:sin(x/2)>0,即2kπ
sin2(x-π/4)=sin(2x-π/2)=-cos2x=-(1-2(sina)^2)=2(sina)^2-1因为sinx=(√5-1)/2,所以sin2(x-π/4)=2((√5-1)/2)^2
(1)f(x)=3sin(2x-π6)+1-cos(2x-π6)=2[32sin(2x-π6)-12cos(2x-π6)]+1=2sin(2x-π3)+1,∵ω=2,∴T=π;(2)令2x-π3=2k
f(x)=sin(x/2)cos(x/2)+√3*sin²(x/2)+√3/2=1/2*sinx+√3/2*(1-cosx)+√3/2=1/2*sinx-√3/2*cosx+√3=sin(x
(I)∵f(x)=2sin2(π4+x)-3cos2x=1-cos(π2+2x)-3cos2x=1+sin2x-3cos2x=2sin(2x-π3)+1.(1分)∴周期T=π;(1分)令2kπ-π2≤