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)=√3⼀3,则cos(a+b⼀2)=

2025-05-22 09:17:21
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回答1:

0-π/2cos(a+b/2)=cos[π/4+a-(π/4-b/2)]=cos(π/4+a)*cos(π/4-b/2)+sin(π/4+a)*sin(π/4-b/2)=5√3/9