sin4θ+cos4θ=1 sinθ+cosθ=？

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

所以cos4θ=1 接下来就很简单了

答案补充
答案为1 正负根号2 0

答案补充
（sin^2θ)^2+(cos^2θ)^2=（1-cos^2θ)^2+(cos^2θ)^2=1
1+2cos^4θ-2cos^2θ=1
然后用t代替cos^2θ 解出cos^2θ为0 1 所以θ=0 正负1 正负根号2

sin4θ+cos4θ=1
2sin2θcos2θ+cos^2 2θ-sin^2 2θ=1
2sin2θcos2θ=2sin^2 2θ

sin2θ=0或cos2θ=sin2θ

2θ=npi或2θ=1/4pi+npi

sinθ+cosθ=sinn/2pi+cosn/2pi=1或-1
或sinθ+cosθ=sin(1/4pi+npi)+cos(1/4pi+npi)=根号2或-根号2

特殊值θ=0 符合sin4θ+cos4θ=1
所以sinθ+cosθ=1