因为0<a<π/2,-π/2<b<0,cos(π/4+a)=1/3,cos(π/4-b/2)=根号3/3,
所以
π/4+a,π/4-b/2为锐角
即sin(π/4+a)=2√2/3,sin(π/4-b/2)=根号6/3
所以
cos(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)
=1/3*√3/3+2√2/3*√6/3
=(√3+4√3)/9
=5√3/9
因为0<a<π/2,-π/2<b<0,cos(π/4+a)=1/3,cos(π/4-b/2)=根号3/3,
所以
π/4+a,π/4-b/2为锐角
即sin(π/4+a)=2√2/3,sin(π/4-b/2)=根号6/3
所以
cos(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)
=1/3*√3/3+2√2/3*√6/3
=(√3+4√3)/9
=5√3/9