Question

已知0《x《π/4，sin（π/4-x）=5/13，求cos2x/cos（π/4+x）

Answer 1

已知0＜x＜π/4,sin(π/4-x)=5/13,求cos2x/cos(π/4+x)的值
已知0＜x＜π/4,则sinx>0,cosx>0
sin(π/4-x)=sinπ/4·cosx-cosπ/4·sinx=√2/2(cosx-sinx)=5/13
则cosx-sinx=5√2/13，又因为sin²x+cos²x=1
2cosx·sinx=-(cosx-sinx)²+sin²x+cos²x=1-50/169=119/169
(cosx+sinx)²=(cosx-sinx)²+4cosx·sinx=50/169+238/169=288/169
则cosx+sinx=12√2/13
cos2x/cos(π/4+x)=(cos²x-sin²x)/(cosπ/4·cosx-sinπ/4·sinx)=(cosx+sinx)(cosx-sinx)/[√2/2(cosx-sinx)]=√2(cosx+sinx)=√2×12√2/13=24/13

Answer 2

sin（π/4-x）=5/13 , √2/2(cosx-sinx)=5/13 , (cosx-sinx)^2=50/169 , (1-sin2x)=50/169
sin2x=119/169 , cos2x=120/169 , sin（π/4-x）=sin[π/2-（π/4+x）]=cos（π/4+x）=5/13
cos2x/cos（π/4+x）=(120/169)/(5/13)=24/13