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sinx%Cosx=1/2 求sin^3x%Cos^3x的值

sinx-cosx=½ sin²x-2sinxcosx+cos²x=¼ ∴sinxcosx=⅜ sin³x-cos³x=(sinx-cosx)(sin²x+sinxcosx+cos²x) =½(1+⅜) =11/16

(sinx-cosx)(sin^2x+cos^2x+sinxcosx)=(sinx-cosx)(1+sinxcosx)=(sinx-cosx)(1+(sin2x)/2)=(sinx-cosx)(2+sin2x)/2 自己算了!~~

解: 立方差公式: a^3-b^3=(a-b)(a^2+ab+b^2) 则: sin^3x-cos^3x =(sinx-cosx)[sin^2(x)+cos^2(x)+sinxcosx] =(sinx-cosx)[1+sinxcosx] =-1 由于: sinxcosx =(1/2)(2sinxcosx) =(1/2){[sin^2(x)+cos^2(x)]-[sinx-cosx]^2} =(1/2)[1-(sinx-cosx)^2...

把sinx和cosx分别看做a和b,题变为a^3+b^3

设sinx+cosx=t sinxcosx=[(sinx+cosx)??-1]/2=(t??-1)/2 sin??x+cos??x =(sinx+cosx)(sin??x+sinxcosx+cos??x) =t[1+(t??-1)/2]=1 即t??+t-2=0 (t-1)(t??+t+2)=0 易知t??+t+2>0 故t=1 即sinx+cosx=1 故sinxcosx=0 sin^4x+cos^4x=(sin??x+co...

1. sinX-cosX=1/5 1-2sinxcosx=1/25 sinxcosx=12/25 2. (sinx+cosx)²=1+2sinxcosx=1+24/25=49/25 因为sinX-cosX=1/5,0

(1)cosx-2sinx=0 解:tanx=1/2 x=kπ +arctan(1/2),k∈Z (2)sinx=cos3x 解:sin(2x-x)=cos(2x+x) sin2xcosx-cos2xsinx=cos2xcosx-sin2xsinx (sin2x-cos2x)(cosx+sinx)=0 tan2x=1 or tanx=-1 x=k(π/2)+π/8, or x=kπ+π/4,k∈Z (3)sin^2x-3cos^2...

利用积化和差公式: sinXsin2Xsin3X = -(1/2)(cos3X-cosX)sin3X 满意请采纳,谢谢!

cosx-cos3x=cosx-[cos2xcosx-sin2xsinx] =cosx-cos2xcosx+sin2xsinx =cosx(1-cos2x)+sin2xsinx =cosx*2sin^2x+sin2xsinx =sin2xsinx+sin2xsinx =2sin2xsinx

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