Question

已知cos(π/4+x)=4/5,且3π/2<x<7π/4,则sinx等于

Answer 1

∵3π/2<x<7π/4,
∴7π/4<π/4+x<2π,
∵cos(π/4+x)=4/5,
∴sin(π/4+x)=-√[1-cos²(π/4+x)]=-3/5

∴sinx=sin[(π/4+x)-π/4]
=sin(π/4+x)cosπ/4-cos(π/4+x)sinπ/4
=-3/5*√2/2-4/5*√2/2=-7√2/10

Answer 2

因为3π/2<x<7π/4,
所以7π/4<π/4+x<2π,
因为cos(π/4+x)=4/5,
所以sin(π/4+x)=-√[1-cos²(π/4+x)]=-3/5

所以sinx=sin[(π/4+x)-π/4]
=sin(π/4+x)cosπ/4-cos(π/4+x)sinπ/4
=-3/5*√2/2-4/5*√2/2=-7√2/10