Question

在三角形ABC中,AC=2,BC=1,cosC=3/4,AB求出为根号2.求sin(2A+C)的值

Answer 1

b=2,a=1,c=√2

cosC=3/4
sin²C+cos²C=1
0<C<180则sinC>0
sinC=√7/4
a/sinA=b/sinB=c/sinC
1/sinA=2/sinB=√2/(√7/4)
sinA=√14/8
sinB=√14/4
sin²A+cos²A=1
cosA=5√2/8
cosB=(a²+c²-b²)/2ac=√2/4

2A+C=A+(A+C)=A+180-B=180-(B-A)
所以sin(2A+C)=sin(B-A)
=sinBcosA-cosBsinA
=

Answer 2

(1+1/n)+(1+2/n)+(1+3/n)+¨¨¨+[1+(n-1/n)]
=1+1+……+1+1/n+2/n+……+(n-1)/n
=(n-1)+[1+2+……+(n-1)]/n
=(n-1)+[n(n-1)/2]/n
=(n-1)+(n-1)/2
=(3/2)n-3/2

所以原式=lim(n→∞)(1/n)[(3/2)n-3/2]
=lim(n→∞)[3/2-3/(2n)]
=3/2-0
=3/2

Answer 3

倾斜角是a
则k=tana=(1+3)/(2-m)=4/(2-m)

m<2,则k>0
所以a=arctan[4/(2-m)]

m>2,k<0
a=π+arctan[4/(2-m)]

综上
m<2，a=arctan[4/(2-m)]
m>2，a=π+arctan[4/(2-m)]