求sinx∧3+cosx∧3和sinx∧4+cosx∧4的值,
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求sinx∧3+cosx∧3和sinx∧4+cosx∧4的值,
![](http://img.wesiedu.com/upload/4/0d/40d40566046a67b13ff90673bdf77881.jpg)
![](http://img.wesiedu.com/upload/4/0d/40d40566046a67b13ff90673bdf77881.jpg)
![求sinx∧3+cosx∧3和sinx∧4+cosx∧4的值,](/uploads/image/z/6593053-13-3.jpg?t=%E6%B1%82sinx%E2%88%A73%2Bcosx%E2%88%A73%E5%92%8Csinx%E2%88%A74%2Bcosx%E2%88%A74%E7%9A%84%E5%80%BC%2C)
sinx+cosx=m
则(sinx+cosx)^2=m^2
===> 1+2sinxcosx=m^2
===> sinxcosx=(m^2-1)/2
sin^3 x+cos^3 x=(sinx+cosx)(sin^2 x-sinxcosx+cos^2 x)
=m*(1-sinxcosx)
=m*[1-(m^2-1)/2]
=m*(1-m^2)/2
sin^4 x+cos^4 x=(sin^2 x+cos^2 x)^2-2sin^2 xcos^2 x
=1-2(sinxcosx)^2
=1-2*[(m^2-1)/2]^2
=(-m^4+2m^2+1)/2
则(sinx+cosx)^2=m^2
===> 1+2sinxcosx=m^2
===> sinxcosx=(m^2-1)/2
sin^3 x+cos^3 x=(sinx+cosx)(sin^2 x-sinxcosx+cos^2 x)
=m*(1-sinxcosx)
=m*[1-(m^2-1)/2]
=m*(1-m^2)/2
sin^4 x+cos^4 x=(sin^2 x+cos^2 x)^2-2sin^2 xcos^2 x
=1-2(sinxcosx)^2
=1-2*[(m^2-1)/2]^2
=(-m^4+2m^2+1)/2
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