1-cos+2x等于多少
@别婵3553：高一数学求解:1 - 2sin^2X等于多少 -
充蚀13337241897…… 等于cos^2X 1-sin^2X=sin^2X+cos^2X-sin^2X=cos^2X 求采纳
@别婵3553：1+cos2x=2cos^x 怎么推导? -
充蚀13337241897…… 1+cos2x=1+cos(x+x)=cosxcosx-sinxsinx+1=cos^x+1-sin^x=cos^x+cosx^x=2cos^x
@别婵3553：大佬们1+cos^2x等于多少?
充蚀13337241897…… 你好:1+cos²x=sin²x+cos²x+cos²x=sin²x+2cos²x=sin²x+cos2x+1
@别婵3553：(1+cos x/2)/2等于多少? -
充蚀13337241897…… =1/2+cosX
@别婵3553：急!!f(x)=sin^2x+sin^2 (x+α)+sin^2 (x+β)若fx是一个与x无关的常值函数,求a,b的值 -
充蚀13337241897…… f(x)=sin²x+sin²(x+α)+sin²(x+β)=(1-cos2x)/2 +(1-cos(2x+2α))/2 +(1-cos(2x+2β))/2=3/2 -[cos2x+cos(2x+2α)+cos(2x+2β)]/2=3/2-(cos2x+cos2xcos2α-sin2xsin2α+cos2xcos2β-sin2xsin2β)/2=3/2 -[(1+cos2α+cos2β)cos2x -(sin2α+sin2β)sin2x]/2 如果使...
@别婵3553：求方程式:1+COS x + COX 2x =0 , 0<=x<=360度 的解 -
充蚀13337241897…… 1+cosx+ cos2x =01+cosx+2cos^2x -1=0 cosx+2cos^2x=0 cosx(1+2cosx)=0 cosx=0或1+2cosx=0 cosx=0或cosx=-1/2 x=90度或x=270度或x=120度或x=240度
@别婵3553：函数f(x)=cos^2(x - π/12)+sin^2(x+π/12) - 1是周期为什么的奇还是偶函数 -
充蚀13337241897…… cos^2(x-π/12)=[1+cos(2x-π/6)]/2 sin^2(x+π/12)=[1-cos(2x+π/6)]/2 所以f(x)=[cos(2x-π/6)-cos(2x+π/6)]/2=(cos2xcosπ/6+sin2xsinπ/6-cos2xcosπ/6+sin2xsinπ/6)/2=sin2xsinπ/6 所以T=2π/2=π f(x)=(sin2x)/2,f(-x)=[sin(-2x)]/2=-(sin2x)/2 定义域是R,关于原点对称 所以是奇函数
@别婵3553：f(sinx)=1+cos(2x),求f(x),详细解答 -
充蚀13337241897…… cos2X=(cosX)^2-(sinX)^2 =2*(cosX)^2-1 =1-2*(sinX)^2 f(x)=1-2x^2
@别婵3553：f(x)=(1+cos 2x)*sin^2 x -
充蚀13337241897…… f(x)=(1+cos 2x)*sin^2 x =(1/2)(1+cos2x)(1-cos2x) =(1/2)(1-cos^2 2x) =(1/2)sin^2 x =(1/4)(1-cos2x) 最小正周期=2pi/2=pi 是偶函数
@别婵3553：f(sin x/2)=1+cos x那么f(cos x/2)= -
充蚀13337241897…… 1+cosx=2(cos(x/2))^2=2-2(sin(x/2))^2 f(sin x/2)=1+cos x=2-2(sin(x/2))^2 因此f(t)=2-2t^2 f(cosx/2)=2-2(cos(x/2))^2=2(sin(x/2))^2 =1-cosx
充蚀13337241897…… 等于cos^2X 1-sin^2X=sin^2X+cos^2X-sin^2X=cos^2X 求采纳
@别婵3553：1+cos2x=2cos^x 怎么推导? -
充蚀13337241897…… 1+cos2x=1+cos(x+x)=cosxcosx-sinxsinx+1=cos^x+1-sin^x=cos^x+cosx^x=2cos^x
@别婵3553：大佬们1+cos^2x等于多少?
充蚀13337241897…… 你好:1+cos²x=sin²x+cos²x+cos²x=sin²x+2cos²x=sin²x+cos2x+1
@别婵3553：(1+cos x/2)/2等于多少? -
充蚀13337241897…… =1/2+cosX
@别婵3553：急!!f(x)=sin^2x+sin^2 (x+α)+sin^2 (x+β)若fx是一个与x无关的常值函数,求a,b的值 -
充蚀13337241897…… f(x)=sin²x+sin²(x+α)+sin²(x+β)=(1-cos2x)/2 +(1-cos(2x+2α))/2 +(1-cos(2x+2β))/2=3/2 -[cos2x+cos(2x+2α)+cos(2x+2β)]/2=3/2-(cos2x+cos2xcos2α-sin2xsin2α+cos2xcos2β-sin2xsin2β)/2=3/2 -[(1+cos2α+cos2β)cos2x -(sin2α+sin2β)sin2x]/2 如果使...
@别婵3553：求方程式:1+COS x + COX 2x =0 , 0<=x<=360度 的解 -
充蚀13337241897…… 1+cosx+ cos2x =01+cosx+2cos^2x -1=0 cosx+2cos^2x=0 cosx(1+2cosx)=0 cosx=0或1+2cosx=0 cosx=0或cosx=-1/2 x=90度或x=270度或x=120度或x=240度
@别婵3553：函数f(x)=cos^2(x - π/12)+sin^2(x+π/12) - 1是周期为什么的奇还是偶函数 -
充蚀13337241897…… cos^2(x-π/12)=[1+cos(2x-π/6)]/2 sin^2(x+π/12)=[1-cos(2x+π/6)]/2 所以f(x)=[cos(2x-π/6)-cos(2x+π/6)]/2=(cos2xcosπ/6+sin2xsinπ/6-cos2xcosπ/6+sin2xsinπ/6)/2=sin2xsinπ/6 所以T=2π/2=π f(x)=(sin2x)/2,f(-x)=[sin(-2x)]/2=-(sin2x)/2 定义域是R,关于原点对称 所以是奇函数
@别婵3553：f(sinx)=1+cos(2x),求f(x),详细解答 -
充蚀13337241897…… cos2X=(cosX)^2-(sinX)^2 =2*(cosX)^2-1 =1-2*(sinX)^2 f(x)=1-2x^2
@别婵3553：f(x)=(1+cos 2x)*sin^2 x -
充蚀13337241897…… f(x)=(1+cos 2x)*sin^2 x =(1/2)(1+cos2x)(1-cos2x) =(1/2)(1-cos^2 2x) =(1/2)sin^2 x =(1/4)(1-cos2x) 最小正周期=2pi/2=pi 是偶函数
@别婵3553：f(sin x/2)=1+cos x那么f(cos x/2)= -
充蚀13337241897…… 1+cosx=2(cos(x/2))^2=2-2(sin(x/2))^2 f(sin x/2)=1+cos x=2-2(sin(x/2))^2 因此f(t)=2-2t^2 f(cosx/2)=2-2(cos(x/2))^2=2(sin(x/2))^2 =1-cosx
