sin^4θ+cos^4θ=1求sinθ-cosθ=多少?

乍看一眼从4次方变成没次方很有难度,但如果利用(sinθ)^2+(cosθ)^2=1的话就能降次数了.（ps：把sin^4θ写成(sinθ)^4会比较好,否则会让人跟sin4θ搞不清）
θ是第三象限角,所以sinθ

sin^4α+cos^4α=1
sin^4α+cos^4α+2sin^2α*cos^2α-2cos^2α*sin^2α=1
(sin^2α+cos^2α)^2-2cos^2α*sin^2α=1
cosαsinα=0
所以有 cosα=0 或者 sinα=0 也就是 α=kπ/2
cosα=0 时,sinα=±1
sinα=0 时,cosα=±1
所以
sin^kα+cos^kα= (±1)^k

sin^4a＋cos^4a
=(sin²a＋cos²a)²－2sin²acos²a
=1－(1/2)sin²2a
=1－(1/2)[1－cos²2a]
=11/18

证明：右柯西不等式可知
1=sin^4 α/cos^2 β+cos^4 α/sin^2 β
=(cos^2β+sin^2β)(sin^4 α/cos^2 β+cos^4 α/sin^2 β)
≥(sin^2α+cos^2α)^2=1
当且仅当（sin^4 α/cos^2 β)/(cos^2β)=(cos^4 α/sin^2 β)/(sin^2β)时等号成立,即
sin^2α/cos^2α=cos^2 β/sin^2 β
所以sin^4 β/cos^2 α+cos^4 β/sin^2 α
=cos^2βsin^2β/sin^2α+cos^2βsin^2β/cos^2 α
=cos^2βsin^2β(1/sin^2α+1/cos^2 α)
=cos^2βsin^2β/(sin^2αcos^2 α)
=1
所以sin^4 β/cos^2 α+cos^4 β/sin^2 α=1

sin2α=（根号2）/2
sin^4α+cos^4α=(sin^2a+cos^2a)^2-2sin^2acos^2a=1-1/2 sin^2(2a)= 1 - 1/2*(√2/2)^2 = 3/4

（1）sin^2·α+cos^4·α+sin^2·αcos^2·α
=sin^2·α+cos^2·α(cos^2·α+sin^2·α)
= sin^2·α+cos^2·α
=1
（2）（1-cos^4·α-sin^4·α）/(1-cos^6·α-sin^4·α)
=[(1+sin^2·α)(1-sin^2·α)-cos^4·α]/[(1+sin^2·α)(1-sin^2·α)-cos^6·α]
=[cos^2·α+cos^2·αsin^2·α-cos^4·α]/[cos^2·α+cos^2·αsin^2·α-cos^6·α]
=[cos^2·α(1-cos^2·α)+cos^2·αsin^2·α]/[cos^2·α(1-cos^4·α)+cos^2·αsin^2·α]
=[2cos^2·αsin^2·α]/[cos^2·α(1-cos^2·α)(1+cos^2·α)+cos^2·αsin^2·α]
=[2cos^2·αsin^2·α]/[cos^2·αsin^2·α(1+cos^2·α)+cos^2·αsin^2·α]
=[2cos^2·αsin^2·α]/[2cos^2·αsin^2·α+sin^2·αcos^4·α]
=2/[2+cos^2·α]

sinθ+sin^2θ=1,sin^2θ+cos^2θ=1
∴sinθ=cos^2θ
等式两边同时乘以sinθ有
sin^2θ+sin^3θ=sinθ
∴sin^3θ=sinθ-sin^2θ=cos^2θ-sin^2θ=cos2θ
∴
=sinθ+sin^2θ+sin^3θ
=1+sin^3θ
=1+cos2θ
=2cos^2θ
=2sinθ
根据sin²θ+sinθ-1=0
解出sinθ=（-1±根号5）/2,带入就行了

分子=(sinx)^6+(cosx)^6=(sin²x)³+(cos²x)³=(sin²x+cos²x)[(sinx)^4-sin²xcos²x+(cosx)^4]=(sin²x+cos²x)²-3sin²xcos²x=1-3(sinxcosx)².分母=（sin²x+cos²x)²-2(sinxcosx)²=1-2(sinxcosx)².∴原式=[1-3(sinxcosx)²]/[1-2(sinxcosx)²].只能这样了,求不出值的.请再看看题.

sinα-cosα=m 1-2sinacosa=m^2 2sinacosa=1-m^2
sin^4α+cos^4α=(sin^2α+cos^2α)-2sin^2acos^2a=1-1/2(2sinacosa)^2
=1-1/2(1-2m^2+m^4)
=1/2+m^2-1/2m^4

∵（sinα）^4/（cosβ）^2＋（cosα）^4/（sinβ）^2＝1,
∴（sinα）^4（sinβ）^2＋（cosα）^4（cosβ）^2＝（sinβ）^2（cosβ）^2,
∴［1－（cosα）^2］^2（sinβ）^2＋（cosα）^4（cosβ）^2＝（sinβ）^2（cosβ）^2,
∴（sinβ）^2－2（cosα）^2（sinβ）^2＋（cosα）^4（sinβ）^2＋（cosα）^4（cosβ）^2
　＝（sinβ）^2（cosβ）^2,
∴（sinβ）^2－2（cosα）^2（sinβ）^2＋（cosα）^4［（sinβ）^2＋（cosβ）^2］
　＝（sinβ）^2（cosβ）^2,
∴（sinβ）^2［1－（cosβ）^2］－2（cosα）^2（sinβ）^2＋（cosα）^4＝0,
∴（sinβ）^4－2（cosα）^2（sinβ）^2＋（cosα）^4＝0,
∴［（sinβ）^2－（cosα）^2］^2＝0,
∴（sinβ）^2＝（cosα）^2,
∵α、β都是锐角,∴sinβ＞0、cosα＞0,∴sinβ＝cosα,∴α＋β＝π/2.

cos2θ=2(cosθ)^2-1=-3/4
(cosθ)^2=1/8 (sinθ)^2=7/8
sin^4θ+cos^4θ=(7/8)^2+(1/8)^2=25/32
欢迎追问~

1/4sin^2(2α)+sin^2β+cos^4α
=1/4(2sinacosa)^2+sin^2β+cos^4α
=sin^2acos^2a+sin^2β+cos^4α
=cos^2a(sin^a+cos^2a)+sin^2β
=sin^2β+cos^2a
=(1-cos2β)/2+(cos2a-1)/2
=1/2(cos2a-cos2β)
=1/2*(-2)(sin(2a+2β)/2sin(2a-2β)/2
=-sin(a+β)sin(a-β)=-1/3
所以本题得证

用a代替
sina=1-sin²a=cos²a
所以原式=sina+sin²a+sin³a
=sina+sina(sina+sin²a)
=sina+sina
=2sina

sin^4α+cos^4α+2sin^2α*cos^2α-1
=(sin²a+cos²a)²-1²
=1-1
=0

sin^4θ+cos^4θ=sin2θ
(sin²θ + cos²θ)² - 2sin²θcos²θ = sin2θ
1² - 2sinθcosθ * sinθcosθ = sin2θ
1 - sin2θ * 1/2 * 2sinθcosθ = sin2θ
1 - 1/2 * sin² 2θ = sin2θ
2 - sin² 2θ = 2sin2θ
sin² 2θ + 2sin2θ = 2
sin² 2θ + 2sin2θ + 1 = 3
(sin2θ + 1)² = 3
sin2θ = √3 - 1 或 -√3 - 1

sin^4a＋cos^4a
=(sin²a＋cos²a)²－2sin²acos²a
=1－(1/2)sin²2a
=1－(1/2)[1－cos²2a]
=1－(1/2)[1－(2/9)]
=11/18