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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
人气:478 ℃ 时间:2020-02-05 09:50:07
解答
m.n =1(-1,√3).(cosA,sinA)=1-cosA+√3sinA =1-1/2cosA +√3/2sinA = 1/2sin(A-π/6) = 1/2A-π/6 = π/6 or 5π/6A = π/3or π2sinBcosB/((cosB)^2-(sinB)^2) = -3sin2B/co2B = -3tan2B = -3tan2B = 2tanB/...
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