X is the inner angle of the triangle, and sinxcosx = - 1 / 8, find the value of SiNx cos

X is the inner angle of the triangle, and sinxcosx = - 1 / 8, find the value of SiNx cos

sinxcosx=-1/80
Then cosx0
(sinx-cosx)²
=sin²x+cos²x-2sinxcosx
=1-2×(-1/8)
=5/4
sinx-cosx>0
So SiNx cosx = √ 5 / 2