asinx cosx等于

f(x)=2asinx^2-（2√3）asinxcosx+b=-a(1-2sinx^2)-2√3asinxcosx+b+a=-acos2x-a√3sin2x+a+b=-2a(0.5cos2x+0.5√

2sin^2x=1-cos2x2sinxcosx=sin2xf(x)=a-acos2x-a√3sin2x+b=a+b-2a(cos2xcosπ/3+sin2xsinπ/3)=a+b-2acos(2x-

f(x)=2asin^2x-2√2asinx+a+b=2a（sinx-√2/2）^2+b定义域为【0,π/2】,则sinx∈【0,1】1、当a>0时,当sinx=√2/2,取得最小值,f（x）min=

fx=2asin²x-2√3asinxcosx+a+b=a(2-cos2x-√3sin2x)+b=a[2-2sin(2x+π/6)]+b2π/3

已知函数f(x)=2asinxcosx+2bcos²x,f(π/6)=6+(3√3)/2,f(0)=8【求a,b的值和f(x)的周期和最大值】f(x)=2asinxcosx+2bcos

括号中是sinx+cosx吗?f(x)=2asinxcosx-√2a(sinx=cosx)+a+b=[(sinx+cosx)²-1]a-a√2(sinx+cosx)+a+b因为x∈[0,π/

f(x)=2acos²x+2根号下3asincosx-a+b=acos2x+根号下3asin2x+b=2asin(2x+π/6)+bx∈[0,π/2]2x+π/6∈[π/6,7π/6]sin

f(x)=a((sinx)^2+1-(cosx)^2)-√3asin2x+b+1=a(1-cos2x)-√3sin2x+b+1=a-2asin(2x+π/6)+b+1,x∈[0,π/2],∴1/2≤s

f(x)=2asinx^2-2√3asinxcosx+b+a=2asinx^2-2√3/2sin2x+b+a,观察可知,当x=π/2时,sinx=1,sin2x=0,f(x)取最大值=1,即f(π/2

y=2asin^x-2√3asinxcosx+b=a(1-cos2x)-a√3sin2x+b=(b+a)-2a((1/2)cos2x+(√3/2)sin2x)=(b+a)-2asin(2x+π/6)最

1）f平移后的函数为-4cos2(平方)（x+π/4)+4(根号3)asin(x+π/4)cos(x+π/4)-2=-2[1+cos(2x+π/2)]+2(根号3)asin(2x+π/2)-2=-2c

f(x)=2asinx^2-2√3asinxcosx+a+b=a-acos2x-√3asin2x+a+b=-2asin(2x+π/6)+2a+b定义域为[π/2,π],7π/6

（Ⅰ）f(x)＝asinxcosx−2cos2x+1＝a2sin2x−cos2x（3分）依题意得f(π8)＝0，即a2sinπ4−cosπ4＝0，解得a=2（6分）（Ⅱ）由（Ⅰ）得f(x)＝sin2x

f（x)=a/2sin2x-a/2(cos2x+1)+√3a／2+b=a/2(sin2x-cos2x)-a/2+√3a／2+b=√2a/2sin(2x-π／4)-a/2+√3a／2+b1)π／2+2k

f(x)=2√3a*sinxcosx-2asin²x+2a+b+1=√3a*sin2x+a*cos2x+a+b+1=2a*sin(2x+π/6)+a+b+1x∈[0,π/2]所以2x+π/6

2asin^x-2根号3asinxcosx+a+b=2asin²x-√3a(2sinxcosx)+a+b=-(a-2asin²x)+a-√3a(2sinxcosx)+a+b=-ac

∵f(x)=-acos2x-√3asin2x+2a+b=-2a(cos2x/2+√3/2*sin2x)+2a+b=-2asin(2x+π/6)+2a+b∵0≤x≤π/2∴π/6≤2x+π/6≤7π/6

f(x)=asinxcosx-2cos²x+1=(a/2)sin2x-cos2x.(1)由题设知,0=f(π/8)=(a/2)sin(π/4)-cos(π/4).===>a/2=1.===>

f'(x)=2a[cos2x-cos(x+派/4)]令f'(x)=0=〉x=派/4又f（0）=（1-根号2)a+b=f(派/2)=（1-根号2)a+b所以f(派/4)=1f(0)=-5或f(派/4)=

1、a=1f(x)=3-4sinxcosx+4cos²x(1-cos²x)=3-4sinxcosx+4sin²xcos²x=3-2sin2x+sin²