sin4θ+cos4θ怎样化简?4是四次方

sin4θ+cos4θ
=(sin^2θ)^2+(cos^2θ)^2+2(sin^2θ)*cos^2θ)-2(sin^2θ)*cos^2θ)
=[(sin^2θ)+(cos^2θ)]^2-2(sin^2θ)*cos^2θ)
=1-(1/2)*(sin2θ)^2
=1-(1/2)*[(1-cos4θ)/2
=3/4+(1/4)*cos4θ

(sinθ)^2(1-(cosθ)^2)+cosθ^2(1-(sinθ)^2)
=(sinθ)^2+cosθ^2-2(cosθsinθ)^2
=1-1/2(2sinθcosθ)(2sinθcosθ)
=1-1/2 (sin2θ)^2
=1-1/4(2sin2θ)^2-1)-1/4
=1-1/4*sin4θ-1/4
=3/4+sin4θ/4,

本答案利用(sinθ)^2+(cosθ)^2=1解的，再配方
容易理解！容易明白！
(sinθ)^4+(cosθ)^4
=[(sinθ)^2]^2+[(cosθ)^2]^2+2[(sinθ)^2]*[(cosθ)^2]-2[(sinθ)^2]*[(cosθ)^2]
=[(sinθ)^2+(cosθ)^2]^2-2[(sinθ)^2]*[(cosθ)^2]
=1-(sin2θ)^2/2
=1-1/4(2sin2θ)^2-1)-1/4
=1-1/4*sin4θ-1/4
=3/4-sin4θ/4.

(sinθ)^4+(cosθ)^4=[(sinθ)^2]^2+[(cosθ)^2]^2+2(sinθ)^2*(cosθ)^2-2(sinθ)^2*(cosθ)^2=[(sinθ)^2+(cosθ)^2]^2-(2*sinθcosθ)^2/2=1-(sin2θ)^2/2

原式=√2sin(4θ+π/4)
哦,不好意思.没有看清除问题的补充.
如果是按照字面上的意思,我上面的答案就是正确的.
如果按照你原来的意思,那答案应该是：
(sinθ)^2(1-(cosθ)^2)+cosθ^2(1-(sinθ)^2)
=(sinθ)^2+cosθ^2-2(cosθsinθ)^2
=1-1/2(2sinθcosθ)(2sinθcosθ)
=1-1/2 (sin2θ)^2
=1-1/4(2sin2θ)^2-1)-1/4
=1-1/4*sin4θ-1/4
=3/4-sin4θ/4.