15问答网 > 已知cos(a-b⼀2)=-1⼀9 sin(a⼀2-b)=2⼀3 ,且π⼀2<a<π,0<b<π⼀2 求tan(a+b)⼀2

已知cos(a-b⼀2)=-1⼀9 sin(a⼀2-b)=2⼀3 ,且π⼀2<a<π,0<b<π⼀2 求tan(a+b)⼀2

2024-12-01 10:51:44

推荐回答（1个）

回答1：

∵π/2∴π/4∴sin[a-(b/2)]>0,
cos[(a/2)-b]>0.
∵cos[a-(b/2)]=-1/9,
sin[(a/2)-b]=2/3，
∴sin[a-(b/2)]=4√5/9,
cos[(a/2)-b]=√5/3.
∴cos(a＋b）=cos[（a－b/2）-（a/2 －b）]
=cos（a－b/2）cos（a/2－b）-sin（a－b/）sin（a/2－b)
=-(1/9)*√5/3-4√5/9*2/3
=-√5/3
sin(a+b)
=±√[1-cos²[(a+b)]
=±√[1-(-√5/3)^2]
=±2/3
tan(a+b)/2
=[sin(a+b)/2] / [cos(a+b)/2]
=[2sin(a+b)/2 * cos(a+b)/2]/[2cos(a+b)/2 * cos(a+b)/2]
=[sin(a+b)]/[cos(a+b) +1]
=[±2/3]/[-√5/3 +1]
=±2/[3-√5]
=±2(3+√5)/(3+√5)(3-√5)
=±(3+√5)/2