Question

已知cos（π/4+x)=-3/5,3/4π＜x＜5π/4，求(sin2x+2sin²x)/(1-tanx)的值。

Answer 1

解：
cos(π/4+x)
=cosπ/4cosx-sinπ/4sinx
=(√2/2)(cosx-sinx)
=-3/5
cosx-sinx=-3√2/5
(cosx-sinx)²=18/25
1-2sinxcosx=18/25
2sinxcosx=7/25
(sinx+cosx)²=1+2sinxcosx=1+7/25=32/25
又∵3/4π＜x＜5π/4
∴sinx+cosx<0
sinx+cosx=-4√2/5

(sin2x+2sin²x)/(1-tanx)
=(2sinxcos²x+2sin²xcosx)/(cosx-sinx)
=2sinxcosx(cosx+sinx)/(cosx-sinx)
=(7/25)*(-4√2/5)/(-3√2/5)
=28/75