若0<a<π⼀2 -π⼀2<b<0 cos(π⼀4+a)=1⼀3 cos(π⼀4-b⼀2)=根号3⼀3 则cos(a+b⼀2)=?

2025-05-18 10:13:35
推荐回答(1个)
回答1:

∵0∴π/4<α+π/4<3π/4 π/4<π/4-b/2<π/2
sin(π/4+α)=2√2/3 sin(π/4-b/2)=√6/3
∴cos(a+b/2)=cos((α+π/4)-(π/4-b/2))=cos(α+π/4)cos(π/4-b/2)+sin(α+π/4)sin(π/4-b/2)
=1/3×√3/3+2√2/3×√6/3
=5√3/9