Question

三角函数的问题 (21 13:38:6)

Answer 1

(sinx+cosx)^2=sinx^2+cosx^2+2sinxcosx

=1+2sinxcosx

(sinx-cosx)^2=sinx^2+cosx^2-2sinxcosx

=1-2sinxcosx

两式相加:

(sinx+cosx)^2+(sinx-cosx)^2=2

已知－(π/2)＜x＜0即第四象限,sin x+cos x=1/5

1/25+(sinx-cosx)^2=2

(sinx-cosx)^2=49/25

第四象限:sinx0

sinx-cosx=-7/5