Question

已知sinα+2cosα=0，则sin2α+cos2α=

Answer 1

由题意知sinα=-2cosα
所以sin2α+cos2α=2sinαcosα+(cosα)^2-(sinα)^2=-4(cosα)^2+(cosα)^2-4(cosα)^2=-7(cosα)^2
又因为(sinα)^2+(cosα)^2=1
所以(cosα)^2=1/5
所以sin2α+cos2α=-7(cosα)^2=-7/5

Answer 2

sinα+2cosα=0
sinα=-2cosα
sinα和cosα异号，sin(2α)=2sinαcosα sin(2α)<0
sin²α+cos²α=(-2cosα)²+(cosα)²=5cos²α=1
cos²α=1/5
sin²α=1-1/5=4/5
sin²(2α)=(2sinαcosα)²=4sin²αcos²α=4(1/5)(4/5)=16/25
sin(2α)=-4/5
cos(2α)=2cos²α-1=2/5-1=-3/5
sin(2α)+cos(2α)=-4/5-3/5=-7/5