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

2025-05-23 17:55:54
推荐回答(1个)
回答1:

0所以π/2<π/2+a<π,π/4<π/4-b/2<3π/4
又cos(π/2+a)=1/3,cos(π/4-b/2)=根号3/3,
所以sin(π/2+a)=2根2/3,sin(π/4-b/2)=-根6/3

cos(a+b/2)=cos[(π/2+a)-(π/4-b/2)]
=cos(π/2+a)cos(π/4-b/2)+sin(π/2+a)sin(π/4-b/2)
=..........=-根3/3
记得采纳哦