62问答网 > 已知0<b<π⼀4,π⼀4<a<3π⼀4 cos【(π⼀4)-a】=4⼀5 sin【（3π⼀4）+b】=5⼀13 求cosa ,cos(a+b)

已知0<b<π⼀4,π⼀4<a<3π⼀4 cos【(π⼀4)-a】=4⼀5 sin【（3π⼀4）+b】=5⼀13 求cosa ,cos(a+b)

2025-05-21 11:08:33

推荐回答（1个）

回答1：

∵π/4∴-π<π/4-a<0
∵cos【(π/4)-a】=4/5
∴sin(π/4-a)=-3/5

∵0∴3π/4∵sin【（3π/4）+b】=5/13
∴cos(3π/4+b)=-12/13

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

cos(a+b)
=sin[π/2-(a+b)
=sin[(3π/4+b)-(π/4-a)]
=sin(3π/4+b)cos(π/4-a)-cos(3π/4+b)sin(π/4-a)
=5/13*4/5-(-12/13)*(-3/5)
=-16/65