Question

已知-π/2＜x＜0，sinx+cosx=1/5,

已知-π/2＜x＜0，sinx+cosx=1/5,求sin^2(π+x）+cos(（3/2）π+x)的值

Answer 1

sin^2(π+x）=-sinx
cos(（3/2）π+x)=cosx
所以原式=-sinx+cosx

sin^2x+cos^2x=1
sinx+cosx=1/5
（sinx+cosx）2=sin^2x+cos^2x+2sinxcosx=1/25 2sinxcosx=-24/25
（sinx-cosx）2=sin^2x+cos^2x-2sinxcosx=1+24/25=49/25
sinx-cosx=7/5或-7/5
因为-π/2＜x＜0，sinx cosx均为负值，所以sinx-cosx=-7/5 即-sinx+cosx=7/5
所以sin^2(π+x）+cos(（3/2）π+x)=7/5

Answer 2

sinx+cosx=1/5
sin^2(x)+cos^2(x)=1
-π/2＜x＜0
====>sinx=-3/5; cosx=4/5
sin^2(π+x）+cos(（3/2）π+x)
=sin^2(x)+cos((-1/2)π+x)
=sin^2(x)+sinx
=(-3/5)^2-3/5
=-6/25