在√3sin x cos x=2a-3中,
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f(x)=a(cos²x+sinxcosx)+b=a(cos²x-1/2+sinxcosx+1/2)+b=a(cos2x/2+sin2x/2)+b=a根号下2sin(2x+π/4)
sinx/cosx=3sinx=3cosx(sinx)^2+(cosx)^2=110(cosx)^2=1(cosx)^2=1/10sinxcosx=3(cosx)^2=3/10再问:10(cosx)^
f(x)=-a[1-2(sinx^2)]-√3*a*sin2x+2a+b=-a*cos2x-√3*a*sin2x+2a+b=-2a[1/2*cos2x+(√3)/2*sin2x]+2a+b=-2a*s
f(x)=sin^2x+√3sinxcosx=1-cos²x+√3/2sin2x=1-(1+cos2x)/2+√3/2sin2x=1/2+√3/2sin2x-1/2cos2x=1/2+sin
f(x)=-√3sin^2x+sinxcosx=-√3(1-cos2x)/2+sin2x/2=sin2x/2+√3cos2x/2-√3/2=sin(2x+π/3)-√3/2因为X∈[0,π/2],所以
y=2(cosx)^2+2√3sinxcosx=cos2x+1+2√3sinxcosx=cos2x+√3sin2x+1=2[1/2cos2x+√3/2sin2x)+1=2sin(2x+π/6)+1
fx=sin2x-根号3*(1+cos2x)+a+根号3=2sin(2x-60°)+aT=pi,增区间[k*pi-pi/6,k*pi+5pi/12],k属于Z 2.由题意得-5pi/6<
条件中的乘号应该是点乘吧高中数学没涉及叉乘那我就认为是向量OA·向量OB了(1)f(x)=-2asin²x+2√3asinxcosx+a+b=acos2x-a+√3asin2x+a+b=ac
(1)向量点积为Y=2a+b-2a*sin(2X+π/6)第一种情况a>0所以2Kπ+π/2≤2X+π/6≤2Kπ+3π/2时为增函数,(K属于N)解得X在[Kπ+π/6,Kπ+2π/3]上单调递增第
f(x)=√3sinxcosx+cos²x+a=(sin2x)×(√3/2)+(cos2x+1)×(1/2)+a=(cosπ/6)×sin2x+(sinπ/6)×cos2x+1/2+a=si
解题过程如下:
f(x)=-√3sin²x+sinxcosx=-√3(1-cos2x)/2+(sin2x)/2=1/2sin2x+√3/2cos2x-√3/2=sin2xcos(π/3)+cos2xsin(
tanx=3cosx^2=1/(1+tanx^2)=1/10sinxcosx+cos2x=tanx*cosx^2+2cosx^2-1=3/10+2/10-1=-1/2
f(x)=√3sinxcosx-(sinx)^2-3/2=(√3/2)*sin2x-(1-cos2x)/2-3/2=(√3/2)*sin2x+(1/2)*cos2x-2=sin(2x+π/6)-2x∈
f(x)=2a(sinx)^2+2sinxcosx-a=-acos2x+sin2x=√(1+a²)*sin(2x+γ),γ为对应的角也就是sinγ=-a/√(1+a²),cosγ=
已知函数fx=√3sinxcosx+(cos∧2)x+a(1)求fx的最小正周期及单调递减区间(2)若fx在区间[~π/6,π/3]上的最大值与最小值的和为3/2,求a的值.(1)解析:f(x)=√3
f(x)=(1-cos2x)/2+根号3/2sin2x=1/2+sin(2x-π/6)x∈【π/4,π/2】,2x-π/6∈【π/3,5π/6】,sin(2x-π/6)∈【1/2,1]f(x)max=
y=cos²x+2√3*sinxcosx-sin²x+a=cos²x-sin²x+2√3*sinxcosx+a=cos2x+√3sin2x+a=2(1/2cos
f(x)=2√3sinxcosx+2cos²x+m=√3sin2x+1+cos2x+m=2sin(2x+π/6)+m+1.0再问:在三角形ABC中角ABC所对的边长abc若F(A)=1,si