已知-π/2<x<0,sinx+cosx=1/5 求3sinx^2-2sinxcosx+cosx^2
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已知-π/2<x<0,sinx+cosx=1/5 求3sinx^2-2sinxcosx+cosx^2
![已知-π/2<x<0,sinx+cosx=1/5 求3sinx^2-2sinxcosx+cosx^2](/uploads/image/z/16800767-71-7.jpg?t=%E5%B7%B2%E7%9F%A5-%CF%80%2F2%EF%BC%9Cx%EF%BC%9C0%2Csinx%2Bcosx%3D1%2F5+%E6%B1%823sinx%5E2-2sinxcosx%2Bcosx%5E2)
-π/2<x<0,
(sinx+cosx)^2=1+2sinxcosx=1/25
2sinxcosx=-24/25
(sinx-cosx)^2=1-2sinxcosx=49/25
sinx-cosx=±7/5 ∵sinx+cosx=1/5 ,-π/2<x<0,cosx>0
cosx=4/5
3sinx^2-2sinxcosx+cosx^2
=3-2cosx^2 -2sinxcosx
=3-32/25+24/25
=67/25
(sinx+cosx)^2=1+2sinxcosx=1/25
2sinxcosx=-24/25
(sinx-cosx)^2=1-2sinxcosx=49/25
sinx-cosx=±7/5 ∵sinx+cosx=1/5 ,-π/2<x<0,cosx>0
cosx=4/5
3sinx^2-2sinxcosx+cosx^2
=3-2cosx^2 -2sinxcosx
=3-32/25+24/25
=67/25
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