y=sin2x sin(2x π 3)
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∵cos5π6=-cosπ6∴f(x)=sin2xsinπ6−cos2xcos5π6=sin2xsinπ6+cos2xcosπ6=cos(2x−π6)令-π+2kπ≤2x−π6≤2kπ,得-5π12+
(1)∵函数f(x)=12sin2xsinφ+cos2xcosφ-12sin(π2+φ)(0<φ<π),∴f(x)=12sin2xsin∅+1+cos2x2•cos∅-12cos∅=12s
设x-y=mx+y=n则6m-2n=143m+n=5解得m=2n=-1即x-y=2x+y=-1解得x=1/2y=-3/2
由(1)得3x+3y+2x-2y=365x+y=36(3)由(2)得4x+4y-3x+3y=-20x+7y=-20(4)(3)×7-(4)得34x=272∴x=8把x=8代入(3)得y=-4∴x=8y
{(x+y)/2+(x-y)/3=63(x+y)+2(x-y)=36(1)4(x+y)-3(x-y)=-20(2)由(1)*3+(2)*2得9(x+y)+6(x-y)+8(x+y)-6(x-y)=36
化简得:-x+7y=11①7x+3y=27②①式×7得:-7x+49y=77③②+③得:52y=104∴y=2代入①得:x=3∴x=3,y=2再问:亲,是代入法哦!再答:代入法①式得3x+3y-4x+
(I)f(x)=(2cos2x-1)cosφ+sin2xsinφ=cos2xcosφ+sin2xsinφ=cos(2x-φ),∵f(x)图象过点(π12,1)∴f(π12)=cos(π6-φ)=1,∵
X+Y分之X-Y等于3x=-2yX+Y分之2(x-y)减X+Y分之3X+Y=(-x-3y)/(x+y)=1
因为(x-y)/(x+y)=3,则(x+y)/(x-y)=1/3则5(x-y)(x+y)-(x+y)/2(x-y)=5*3-1/(3*2)=15-1/6=89/6
(1)f(x)=12sin2xsin∅+1+cos2x2cos∅− 12cos∅=12cos(2x−∅)(0<∅<π)∴2×π6−∅=kπ∴∅=π3(2)f(A)=12cos(2A−π3)=
1、f(x)=1/2sin2xsinφ+(1+cos2x)/2cosφ-1/2cosφ=1/2sin2xsinφ+1/2cos2xcosφ=1/2cos(2x-φ)f(π/6)=1/2cos(π/3-
第二个方程是不是写错了2/(x+y)+3/(x-y)=6是这样吗
(1)原式f(x)=1/2sin2xsinφ+cos^xcosφ-1/2cosφ=1/2sin2xsinφ+cosφ(cos^x-1/2)=1/2sin2xsinφ+1/2cos2xcosφ=1/2c
@是α.. f(π/6)=√3/4sinα+3/4cosα-1/2cosα=√3/4sinα+1/4cosα=1/2(√3/2sinα+1/2cosα)=1/2sin(α+π/6)=1/2s
解(x-y)(x+y)-(x-2y)²+x(3x-5y)-(x-y)(x-2y)=(x²-y²)-(x²-4xy+4y²)+(3x²-5xy
(Ⅰ)f(x)=12sin2xsinφ+cos2xcosφ-sin(π2+φ)=12sin2xsinφ+12(1+cos2x)cosφ-12cosφ=12sin2xsinφ+12cos2xcosφ=1
f(x)=1/2sin2xsinφ+cos^2xcosφ-1/2sin(π/2+φ),=1/2sin2xsinφ+1/2(1+cos2x)cosφ-1/2sin(π/2+φ),=1/2sin2xsin
∵cos5π6=-cosπ6∴f(x)=sin2xsinπ6−cos2xcos5π6=sin2xsinπ6+cos2xcosπ6=cos(2x−π6)令-π+2kπ≤2x−π6≤2kπ,得-5π12+