已知cos2θ=3/5,求sin⁴θ+cos⁴θ

已知cos2θ=3/5,求sin⁴θ+cos⁴θ
已知cos2θ=3/5,求sin⁴θ+cos⁴θ

已知cos2θ=3/5,求sin⁴θ+cos⁴θ
答：
已知cos2θ=3/5,求sin⁴θ+cos⁴θ
sin⁴θ+cos⁴θ
=(sin²θ+cos²θ)²-2sin²θcos²θ
=1-(1/2)sin²2θ
=1-(1/2)(1-cos²2θ)
=1-1/2+(1/2)*(3/5)²
=1/2+(1/2)*(9/25)
=17/25

sin^2 a =(1-cos2a)/2=(1-3/5)/2=1/5
cos^2 a=(1+cos2a)/2=(1+3/5)=4/5
所以sin^4 a=1/25 cos^4 a=16/25
sin^4 a+cos^4 a=1/25+16/25=17/25

由cos2θ=2(cosθ)^2-1知，得(cosθ)^2=(cos2θ+1)/2，代入代数式(sinθ)^4+(cosθ)^4=[(sinθ)^2+(cosθ)^2]^2-2(sinθ)^2(cosθ)^2=1-2(cosθ)^2[1-(cosθ)^2]=1-2(cosθ)^2+2(cosθ)^4=1-（cos2θ+1）+[(cos2θ)^2+2cos2θ+1]/2=[(cos2θ)^2+1]/2=[(3/5)^2+1]/2=17/25