Question

sinα+cosα=根号2则sin^4α-cos^4α=?？

Answer 1

sinx+cosx=√2.两边平方
(sinx+cosx)^2=2
(sinx)^2+(cosx)^2+2sinxcosx=2
1+sin2x=2.
sin2x=1
cos2x=0
sin^4α-cos^4α
=(sin^2α-cos^2α)(sin^2α+cos^2α)
=sin^2α-cos^2α
=(sinα-cosα)(sinx+cosx)
=0*√2
=0,2,sin^4α-cos^4α=（sin²α+cos²α）（sin²α-cos²α）
=sin²α-cos²α
=-cos2α
∵sinα+cosα=根号2
∴sin²α+cos²α+2sinαcosα=2
1+sin2α=2
∴sin2α=1 则cos2α=0
则sin^4α-cos^4α=-cos2α=0
【望采纳！谢谢】,2,∵sina+cosa=√2
两边平方,可得
1+sin(2a)=2
∴sin(2a)=1
∴cos(2a)=0
即cos²a-sin²a=0
∴原式=(sin²a+cos²a)(sin²a-cos²a)=0,1,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 (3/,0,