已知cosα=3⼀5,(3π⼀2)<α<2π,求cos(α-π⼀6)及tan2α

已知cosα=3/5,(3π/2)&lt;α&lt;2π,求cos(α-π/6)及tan2α
2025-03-06 03:49:26
推荐回答(2个)
回答1:

cosα=3/5,(3π/2)<α<2π,所以sina=-4/5,tana=-4/3,所以

cos(α-π/6)=cosacosπ/6+sinasinπ/6=3/5*√3/2+(-4/5)*1/2=(3√3-4)/10
tan2a=2tana/(1-tan^2a)=(-8/3)/(1-16/9)=24/7
请采纳,谢谢

回答2:

已知cosα=3/5,(3π/2)<α<2π,求cos(α-π/6)及tan2α
sinα=-4/5
cos(α-π/6)=cosαcos(π/6)+sinαsin(π/6)=(-4+3√3)/5
tanα=-4/3
tan2α=2tanα/(1-tanα*tanα)=3/2=1.5