求值域 y=3-4cos(2x 4 x)-搜答案-神马作文网

由题知,y*(cosx+3)=cosx-3则：cosx=-3(y+1)/(y-1)由cosx的取值范围知：-1≤cosx≤1所以-1≤-3(y+1)/(y-1)≤1由-3(y+1)/(y-1)≥-1解

解题思路：考查三角恒等变换解题过程：varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/read

y=sin²x+cos8x+2sinxcosx+2cos²x=1+sin2x+(1+cos2x)=sin2x+cos2x+2=√2sin(2x+π/4)+2-1

换元法,用t=x方换掉,然后配方.t的范围是大于零.y的值域就出来了

y=2cos²x+5sinx-4y=2(1-sin^2x)+5sinx-4=-2sin^2x+5sinx-2=-2(sinx-5/4)^2+9/8值域当sinx=1的时候有最大值y=1当si

y=(cosx+2)/(sinx-1)ysinx-y=cosx+2ysinx-cosx=y+2√(y²+1)sin(x-t)=y+2,t=arctan(1/y)sin(x-t)=(y+2)/

用两角和差公式拆开化简2cos(x+π/4)cos(x-π/4)=cosxcosx-sinxsinx(这个其实是二倍角公式)=cos2xy=2(0.5cos2x+0.5√3sin2x)=2(sinπ/

y=√2(√2/2*sinα+√2/2cosα)=√2(sinαcosπ/4+cosαsinπ/4)=√2sin(α+π/4)-1

设t=cosx则y=t²+3t+2∵x属于（3π/4,π）∴t∈(-1,-√2/2)∴y=t²+3t+2=(t+3/2)²-1/4在(-1,-√2/2)上单调递增∴t=-

sinx∈[-1,1]cos(sinx)∈[cos1,1]通过余弦函数的图象的对称性可知当sinx=0时cosx最大为1最小为cos(-1)=cos1

令t=sinx,则-1=

令t=sinx+cosx则t=√2sin(x+45°)∈[-√2,√2]而sinxcosx=[(sinx+cosx)^2-(sinx)^2-(cosx)^2]/2=(t^2-1)/2∴原式=t+(t^

（1）y=4-3sin(x-π/3)sinx的值域为[-1,1],所以,4-3sin(x-π/3)值域为[1,7]（2）y=cos^2x-sinx=1-2sin^2x-sinx=-2(sin^2x+1

(cosx)^2=1-(sinx)^2,然后转化为2次函数区间来求

1.y=1/(x2+2x)x²+2x=x²+2x+1-1=(x+1)²-1>=-1令a=(x+1)²-1y=1/a若-1

y=cos^2x-2sinx=1-sin^2x-2sinx=-(sinx+1)^2+2x∈[π/6,π/4],则1/2

因为-2π/3<x<π,所以-2π/3<(1/2)x-π/3<π/6.所以-1/2<cos(1/2x-π/3)<=1.-√2/2<√2cos(1/2x-π/3

令t=cosx,则|t|

y=2sinx+cos^2x=2sinx+1-sin²x=-(sinx-1)²+2已知x∈[π/6,2π/3),那么：sinx∈[1/2,1]所以当sinx=1即x=π/2时,函数