作业帮cos2x等于什么
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/04 17:57:04
(1)根据图信息可知sina=4/5cosa=3/5∴(sina^2+sin2a)/(cosa^2+cos2a)=(sina^2+2sinacosa)/(cosa^2+cosa^2-sina^2)=(
cos2X=(cosX)^2-(sinX)^2=2*(cosX)^2-1
再答：一定等于
f(sinx)=cos2x=1-2sin²x∴f(t)=1-2t²∴f(cosx)=1-2cos²x=-cos2x
sin^2x+cos^2x=1sin^2x+9sin^2x=1sin^2x=1/10Cos2X+Sin2X=cos^2x-sin^2x+2sinxcosx=9sin^2x-sin^2x+6sin^2x
1/2*sin2x再问：过程？
(sinx/cosx+cosx/sinx)cos^2x-sin^2x=sinx*cosx+cos^3x/sinx-sin^3x/cosx-cosx*sin=(cos^4x-sin^4x)/(sinx*
f(sinx(π/2-x))=cos(π-2x)f(cosx)=-cos(2x)
题目条件不充分啊cos2x+sin2x=1cos2x=cos^2x-sin^2xsin2x=2sinxcosxcos2x+sin2x=cos^2x-sin^2x+2sinxcosx
f(sinx)=cos2x=1-2(sinx)^2所以f(x)=1-2x^2f(1/3)=1-2*(1/3)^2=7/9
cos(A+B)=cosAcosB-sinAsinB(1)cos(A-B)=cosAcosB+sinAsinB(2)(2)-(1)cos(A-B)+cos(A+B)=2cosAcosBA+B=3x/2
sinx/cosx=tanx所以sin2x/cos2x=tan2x
sin2x＋cos2x＝√2（√2＆＃47；2sin2x＋√2＆＃47；2＊cos2x）＝√2（cospai＆＃47；4sin2x＋sinpai＆＃47；4cos2x）＝√2sin（2x＋pai＆＃4
1/2sin4x
cos2x=cos(x+x)=cosx*cosx-sinxsinx=cos²x-sin²x=cos²x-(1-cos²x)=2（cosx的平方）-1
f(sinx)=1+1-cos2x=1+2sin²xf(cosx)=1+2cos²x=cos2x+2再问：1-cos2x=???再答：cos2x=cos²x-sin&su
sinx^2+cosx^2=1,这就相当于一个公式,中间的变数x当然可以换成任何值了!
因为sin^2(X)+cos^2(X)=1所以原式=1-cos^2（x）-cos^2（x）=1-2cos^2(x)=-(2cos^2(x)-1)=-cos2x
limx0时1-cos2x/sinx=limx0时[（2x）²/2]/x=0