函数y=cos^4-sin^4x的周期是

y=(sinx)^4+(cosx)^4=[(sinx)^2+(cosx)^2]^2-2(sinx)^2(cosx)^2=1-(1/2)[sin(2x)]^2=1-(1/4)2[sin(2x)]^2=1

y=sin^4x+cos^4x=[(sinx)^2+(cosx)^2]^2-2(sinxcosx)^2=1-2(sinxcosx)^2=1-[(sin2x)^2]/2=1-[1-(cos4x)]/4=

y=sin^4x+cos^4x+1=[(sinx)^2+(cosx)^2]^2-2(sinxcosx)^2+1=1-2(sinxcosx)^2+1=2-[(sin2x)^2]/2=2-[1-(cos4

y=sin^4x+cos^4xy=[(sinx)^2+(cosx)^2]-2(sinxcosx)^2=1-(2sinxcosx)^2/2=1-(sin2x)^2/2=0.75+(1-2(sin2x)^

y=sin^4x+cos^4x=sin^4x+cos^4x+2sin^2xcos^2x-2sin^2xcos^2x=（sin^2x+cos^2x）^2-2sin^2xcos^2x=1-1/2sin^2

y=(sinx)^4+(cosx)^4=[(sinx)^2+(cosx)^2]^2-2(sinx)^2(cosx)^2=1-2(sinx)^2(cosx)^2=1-4(sinx)^2(cosx)^2/

你的表达可能有点问题,是不是［2（sinα）^2＋sin2α］/（1＋tanα）＝k?若是这样,则方法如下：第一个问题：∵π/4＜α＜π/2,∴sinα＋cosα＞0,∴k＝［2（sinα）^2＋2s

二分之派

Y=cos^4x-2sinxcosx-sin^4x=cos^4x-sin^4x-2sinxcosx=(cos^2x-sin^2x)*1-sin2x=cos2x-sin2x=根号2倍的cos(x+4分之

y=sin^4x+cos^4x=[(sinx)^2+(cosx)^2]^2-2(sinxcosx)^2=1-2(sinxcosx)^2=1-[(sin2x)^2]/2=1-[1-(cos4x)]/4=

y=((1+cos2x)/2)^2+((1-cos2x)/2)^2=(1+2cos2x+cos^2(2x))/4+(1-2cos2x+cos^2(2x))/4=[1+cos^2(2x)]/2=1/2+

是求两个函数（1）y=√(sinx)(2)y=√(cosx)的定义域吧还是求(3)y=√sin(cosx)定义域（1）要使y=√(sinx)有意义,须令sinx≥0所以2kπ≤x≤π+2kπ,k∈z即

y=cos⁴x-sin⁴x=(cos²x+sin²x)(cos²x-sin²x)=cos²x-sin²x=cos(2

y=(cos^2x-sin^2x)(cos^2x+sin^2x)=cos2x所以最小正周期为π

/>y=3sinχcosχ－4cos²χ+2(1-cos²χ)=1.5sin2x-6cos²χ+2=1.5sin2x-3(1+cos2x)+2=1.5(sin2x-2co

y=sin^4x-cos^4x=(sin²x+cos²x)(sin²x-cos²x)=sin²x-cos²x=-cos2x,所以当2x=2k

很简单用了两次二倍角公式而已原函数=1-2sin^2xcos^2x=1-1/2X2sinxcosxX2sinxcosx=1-1/2sin^2(2x)Ps.X表示乘号二倍角公式：sin2x=2sinxc

y=sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-2sin^2xcos^2x=1-(1/2)(2sinxcosx)^2=1-0.5sin^2(2x)=

y=cos^4x-sin^4x+2=(cos^2x+sin^2x)(cos^2x-sin^2x)+2=cos^2x-sin^2x)+2=cos2x+2y=cos^4x-sin^4x+2的最小正周期为2

y=7-4sin2x+4cos2x=7+4√2cos(2x+π/4）,y的最小值=7-4√2.