Question

已知-π/2＜x＜0，sinx+cosx+1/5，求列各式的值

(1)sinx*cosx
(2)sinx-cosx

Answer 1

(1)sinx+cosx=1/5
(sinx+cosx)^2=1+2sinxcosx=1/25
2sinxcosx=-24/25
sinxcosx=-12/25
(2)(sinx-cosx)^2=1-2sinxcosx=1-2*(-12/25)=1+24/25=49/25
sinx-cosx=+-7/5
-pai/2<x<0
-1<sinx<0
0<cosx<1
-1<-cosx<0
-2<sinx-cosx<0
sinx-cosx=-7/5

Answer 2

(sinx+cosx)²=cos²x+sin²x+2sinxcosx
sinxcosx=-12/25
(sinx-cosx)²=(sinx+cosx)²-4sinxcosx
sinx-cosx=±7/5
∵-π/2＜x＜0
∴sin-cosx=-7/5