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

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/05 10:07:25

12√2/13
再问：求过程
再答：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 啊，应该是24/13

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