Question

设α是第三象限角,且sinα^4+cosα^4=5/9,求tanα

Answer 1

α是第三象限角，tanα＞0
sinα^4+cosα^4=5/9
(sinα^2+cosα^2)^2=1
sinα^4+cosα^4+2sinα^2cosα^2=1
2sinα^2cosα^2=1-5/9=4/9
sinα^2cosα^2=2/9
sinα^2=tan^2α/(1+tan^2α)
cosα^2=1/(1+tan^2α)
sinα^2cosα^2=2/9
tan^2α/(1+tan^2α) * 1/(1+tan^2α) = 2/9
9tan^2α = 2tan^4α+4tan^2α+2
(2tan^4α-5tan^2α+2)=0
(2tan^2α-1)(tan^2α-2)=0
2tan^2α=1，或tan^2α=2
tanα＞0
tanα=根号2/2，或tanα=根号2

Answer 2

sinα^4+cosα^4=5/9
sinα^4+2sinα^2cos^2α+cosα^4-2sinα^2cos^2α=5/9
(sinα^2+cos^2α)^2-2sinα^2cos^2α=5/9
1-2(1-sinα^2)sin^2α=5/9
1-2sinα^2+2sin^4α=5/9
sinα^4-sinα^2+2/9=0
(sinα^2-1/3)(sinα^2-2/3)=0
sinα^2=1/3,sinα^2=2/3
(1-cos2α)/2=1/3或(1-cos2α)/2=2/3
(1-cos2α)=2/3或(1-cos2α)=4/3
cos2α=1/3或cos2α=-1/3

当cos2α=1/3时
cos2α=(1-tanα^2)/(1+tanα^2)=1/3
(1+tanα^2)=3(1-tanα^2)
1+tanα^2=3-3tanα^2
4tanα^2=2
tanα^2=1/2
tanα=±√2/2 (α是第三象限角)
tanα=√2/2

当cos2α=-1/3时
cos2α=(1-tanα^2)/(1+tanα^2)=-1/3
(1+tanα^2)=-3(1-tanα^2)
1+tanα^2=-3+3tanα^2
2tanα^2=4
tanα^2=2
tanα=±√2 (α是第三象限角)
tanα=√2

综上tanα=√2或tanα=√2/2