Question

已知A,B,C是三角形ABC三个内角,向量m =(-1,根号3),向量n =( cosA, sin A),且向量m 乘以向量 n=1,求角A.

若（1+sin2B）/（cos^B-sin^B）=-3,求tanB,tanC.

Answer 1

(-1,根号3)*( cosA, sin A)=-cosA+√3sinA=1
-cosA+√3sinA
=2sin(A-30°)=1
只有A-30°=30°
A=60°
题目少打了吗？
若（1+sin2B）/（cos^2B-sin^2B）=-3,求tanB,tanC.
（1+sin2B）/（cos^2B-sin^2B）=-3
(cos^2B+sin^2B+2sinBcosB)/（cos^2B-sin^2B）=-3
（cosB+sinB)^2/（cosB-sinB）(cosB+sinB）=-3
（cosB+sinB）/（cosB-sinB）=-3
(1+tanB）/(tanB-1)=-3
tanB=1/2
tanC=-tan(A+B)=-3√3-5
若果题目没错，再重新做

追问

题目没错！请教

追答

（cos^B-sin^B)不是（cos^2B-sin^2B）吗？
那太麻烦了，对不起啊

Answer 2

m.n=1
(-1,√3).( cosA, sin A)=1
-cosA+√3sin A =1
(√3sin A)^2 = (1+cosA)^2
2(cosA)^2 + cosA -1 =0
(2cosA-1)(cosA+1)=0
cosA = 1/2 or -1 (rejected)
A = π/3
(1+sin2B)/(cos^B-sin^B)=-3
(1+ sin2B) /cos2B = -3
sec2B+ tan2B = -3
(3+tan2B)^2 = (sec2B)^2
6tan2B+ 9 = 1
tan2B = -4/3
2tanB/(1-(tanB)^2 ) = -4/3
-4+4(tanB)^2 = 6tanB
2(tanB)^2 -3tanB-2 =0
(2tanB+1)(tanB-2) =0
tanB = 2 or -1/2 (rejected)
ie tanB = 2

tanC
= tan(π-A-B)
= tan(2π/3-B)
= (tan2π/3 - tanB)/(1+tanBtan2π/3)
=(-√3-2)/(1-2√3)
= ( √3+2)(2√3+1)/ [(2√3-1)(2√3+1)]
= (8+5√3)/5

Answer 3

m.n=1
(-1,√3).( cosA, sin A)=1
-cosA+√3sin A =1
(√3sin A)^2 = (1+cosA)^2
2(cosA)^2 + cosA -1 =0
(2cosA-1)(cosA+1)=0
cosA = 1/2 or -1 (rejected)
A = π/3
(1+sin2B)/(cos^B-sin^B)=-3
(1+ sin2B) /cos2B = -3
sec2B+ tan2B = -3
(3+tan2B)^2 = (sec2B)^2
6tan2B+ 9 = 1
tan2B = -4/3
2tanB/(1-(tanB)^2 ) = -4/3
-4+4(tanB)^2 = 6tanB
2(tanB)^2 -3tanB-2 =0
(2tanB+1)(tanB-2) =0
tanB = 2 or -1/2 (rejected)
ie tanB = 2

tanC
= tan(π-A-B)
= tan(2π/3-B)
= (tan2π/3 - tanB)/(1+tanBtan2π/3)
=(-√3-2)/(1-2√3)
= ( √3+2)(2√3+1)/ [(2√3-1)(2√3+1)]
= (8+5√3)/5
嘻嘻