已知5cos(a-b/2)+7cos(b/2)=0,求tan(a/2)*tan((a-b)/2)的值.
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已知5cos(a-b/2)+7cos(b/2)=0,求tan(a/2)*tan((a-b)/2)的值.
(要有具体过程!)
(要有具体过程!)
tan(a/2)*tan[(a-b)/2]
={sin(a/2)/cos(a/2)}×{sin[(a-b)/2]/cos[(a-b)/2]}
={sin(a/2)×sin[(a-b)/2]}/{cos(a/2)×cos[(a-b)/2]}
=(-1/2){cos[(a/2)+(a-b)/2]-cos[(a/2)-(a-b)/2]} / (1/2){cos[(a/2)+(a-b)/2]+cos[(a/2)-(a-b)/2]}
=-{cos[a-(b/2)]-cos(b/2)} / {cos[a-(b/2)]+cos(b/2)}
(提示:根据tanx=sinx/cosx,然后将分子/分母积化和差)
∵5cos[a-(b/2)]+7cos(b/2)=0
∴cos[a-(b/2)]=(-7/5)cos(b/2)
∴tan(a/2)*tan[(a-b)/2]
=-{cos[a-(b/2)]-cos(b/2)} / {cos[a-(b/2)]+cos(b/2)}
=-{(-7/5)cos(b/2)-cos(b/2)} / {(-7/5)cos(b/2)+cos
(b/2)}
=-[(-12/5)cos(b/2)]/[(-2/5)cos(b/2)]
=(12/5)/(-2/5)
=-6
={sin(a/2)/cos(a/2)}×{sin[(a-b)/2]/cos[(a-b)/2]}
={sin(a/2)×sin[(a-b)/2]}/{cos(a/2)×cos[(a-b)/2]}
=(-1/2){cos[(a/2)+(a-b)/2]-cos[(a/2)-(a-b)/2]} / (1/2){cos[(a/2)+(a-b)/2]+cos[(a/2)-(a-b)/2]}
=-{cos[a-(b/2)]-cos(b/2)} / {cos[a-(b/2)]+cos(b/2)}
(提示:根据tanx=sinx/cosx,然后将分子/分母积化和差)
∵5cos[a-(b/2)]+7cos(b/2)=0
∴cos[a-(b/2)]=(-7/5)cos(b/2)
∴tan(a/2)*tan[(a-b)/2]
=-{cos[a-(b/2)]-cos(b/2)} / {cos[a-(b/2)]+cos(b/2)}
=-{(-7/5)cos(b/2)-cos(b/2)} / {(-7/5)cos(b/2)+cos
(b/2)}
=-[(-12/5)cos(b/2)]/[(-2/5)cos(b/2)]
=(12/5)/(-2/5)
=-6
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