Question

已知cos(x–π/4)=√2/10，x∈(π/2，3π/4).(1)求sinx的值(2)求sin(

已知cos(x–π/4)=√2/10，x∈(π/2，3π/4).(1)求sinx的值(2)求sin(2x+π/3)的值

Answer 1

cos(x–π/4)=√2/10
cosxcos(π/4)+sin(x)sin(π/4)=√2/10
cosx+sinx=1/5
平方得cos²x+2cosxsinx+sin²x=1/25
所以cosxsinx=-12/25
所以cosx
，sinx
是t²-(1/5)t-12/25=0
的两个解
4/5,-3/5
x∈(π/2，3π/4)
cosx<0
sinx>0
所以
cosx=-3/5
sinx=4/5
(2)
sin(2x)=2sinxcosx=2(4/5)(-3/5)=-24/25
cos(2x)=1-2sin²x=-7/25
sin(2x+π/3)=sin(2x)cos(π/3)+cos(2x)sin(π/3)
=(-24/25)(1/2)+(-7/25)(√3/2)=
-(24+7√3)/50