高一三角函数题、急

sin[a-(2n+1)π/2]=3/5a属于(0,π/2)并上(π/2,π)求tan(a)
2025-06-23 02:21:24
推荐回答(2个)
回答1:

sin[a-(2n+1)π/2]=-cos(a)=-3/5(奇变偶不变,符号看象限) 则由sin^2(a)+cos^2(a)=1 可求得 sin(a)=-4/5 所以tan(a)=sin(a)/cos(a)=4/3

回答2:

sin[a-(2n+1)π/2]=sina*cos[(2n+1)π/2]-cosa*sin[(2n+1)π/2]=-cosa=3/5
所以cosa=-3/5;sina=-4/5
所以tan(a)=sina/cosa=4/3