若sin^4x+cos^4x=1,则sinθ+cosθ=?

问题描述:

若sin^4x+cos^4x=1,则sinθ+cosθ=?

sin^x+cos^x=1
sinx=√1-cos^x
cox+√1-cos^x

sin^4x+cos^4x=(sin^2x+cos^2x)²-2sin^2xcos^2x=1-2sin^2xcos^2x=1
∴2sin^2xcos^2x =0 (sinxcosx) ²=0
∴sinxcosx =0
(sinx+cosx)²
= sin^2x+cos^2x+2sinxcosx
=1+2sinxcosx
=1
∴sinx+cosx=±1 (当x=kπ/2,k为整数)

sin^4x+cos^4x=(sin^2x+cos^2x)²-2sin^2xcos^2x=1-2sin^2xcos^2x=1∴2sin^2xcos^2x = 0sinxcosx =0(sinx+cosx)²= sin^2x+cos^2x+2sinxcosx=1+2sinxcosx=1∴sinx+cosx=1