sin^3θ+cos^3θ=1,求sinθ+cosθ与sin^4θ+cos^θ的值

解:
(1)
设sinθ+cosθ=x,
则:(sinθ+cosθ)^2=x^2
即:(sinθ)^2+(cosθ)^2+2sinθcosθ=x^2
由于:(sinθ)^2+(cosθ)^2=1
则:sinθcosθ=(x^2-1)/2
由于:(sinθ)^3+(cosθ)^3=1
则有：
(sinθ+cosθ)[(sinθ)^2+(cosθ)^2-sinθcosθ]=1
即:x[1-(x^2-1)/2]=1
x(3-x^2)=2
x^3-3x+2=0
x^3-x^2+(x^2-3x+2)=0
x^2(x-1)+(x-1)(x-2)=0
(x-1)(x^2+x-2)=0
(x-1)(x-1)(x+2)=0
由于：
sinθ+cosθ
=√2sin(θ+∏/4)(辅助角公式)
属于[-√2,√2]
则：x=sinθ+cosθ=1
(2)
(sinθ)^4+(cosθ)^4
=(sinθ)^4+(cosθ)^4+2(sinθcosθ)^2-2(sinθcosθ)^2
=[(sinθ)^2+(cosθ)^2]^2-2(sinθcosθ)^2
=1^2-2*[(x^2-1)/2]^2
=1