已知cos(π⼀4+x)=-3⼀5,3⼀4π<x<5π⼀4,求(sin2x+2sin눀x)⼀(1-tanx)的值。

2024-11-24 16:01:46
推荐回答(1个)
回答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