关于高一数学向量和三角函数的问题已知向量m=（-1,√3）,n=(cosA,sinA),且m×n=1(1)求角A（2）若(2sinB×cosB)/(cos^2 B-sin^2 B)=-3,求tanc
2019-06-25
关于高一数学向量和三角函数的问题
已知向量m=（-1,√3）,n=(cosA,sinA),且m×n=1
(1)求角A
（2）若(2sinB×cosB)/(cos^2 B-sin^2 B)=-3,求tanc
优质解答
m.n =1
(-1,√3).(cosA,sinA)=1
-cosA+√3sinA =1
-1/2cosA +√3/2sinA = 1/2
sin(A-π/6) = 1/2
A-π/6 = π/6 or 5π/6
A = π/3 or π
2sinBcosB/((cosB)^2-(sinB)^2) = -3
sin2B/co2B = -3
tan2B = -3
tan2B = 2tanB/(1-(tanB)^2)
-3=2tanB/(1-(tanB)^2)
-3+3(tanB)^2 = 2tanB
3(tanB)^2 - 2tanB -3 =0
tanB = (1+√10)/3 or (1-√10)/3
m.n =1
(-1,√3).(cosA,sinA)=1
-cosA+√3sinA =1
-1/2cosA +√3/2sinA = 1/2
sin(A-π/6) = 1/2
A-π/6 = π/6 or 5π/6
A = π/3 or π
2sinBcosB/((cosB)^2-(sinB)^2) = -3
sin2B/co2B = -3
tan2B = -3
tan2B = 2tanB/(1-(tanB)^2)
-3=2tanB/(1-(tanB)^2)
-3+3(tanB)^2 = 2tanB
3(tanB)^2 - 2tanB -3 =0
tanB = (1+√10)/3 or (1-√10)/3