已知f(α)=【sin(π-α)*cos(2π-α)]/cos(-π-α)*tan(α+π),求f(-31π/3)的值

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/24 15:21:01
已知f(α)=【sin(π-α)*cos(2π-α)]/cos(-π-α)*tan(α+π),求f(-31π/3)的值
xj@DMS}n>`-*TlTKK!1j!LW5^6E$|Dlt"ܬ3*/ >ﺸs> Lm)l&{@xH?wD?hHίŃcA'f`\+Y&JOte2LA?bT|_9lmBvض6]?XΚqah#xZ6VqY`:Y{4f!2ndTKp7ae;0LjCS4YNz Cr)yHhh*4$jNZԪ@ jV،٩j*ִJΑ$chpHLl!XW^CNS^IrN9 >D.~h[E!;]};7| 

已知f(α)=【sin(π-α)*cos(2π-α)]/cos(-π-α)*tan(α+π),求f(-31π/3)的值
已知f(α)=【sin(π-α)*cos(2π-α)]/cos(-π-α)*tan(α+π),求f(-31π/3)的值

已知f(α)=【sin(π-α)*cos(2π-α)]/cos(-π-α)*tan(α+π),求f(-31π/3)的值
f(α)=【sin(π-α)*cos(2π-α)]/cos(-π-α)*tan(α+π
  =sinαcosα/(-cosα)(-tanα)
  =sinαcosα/sinα
  =cosα
f(-31π/3)=cos(-31π/3)=cos(-10π-π/3)=cosπ/3=1/2
龙者轻吟为您解惑,凤者轻舞闻您追问.
请点击[满意答案];如若您有不满意之处,请指出,我一定改正!
希望还您一个正确答复!
祝您学业进步!


f(α)=[sin(π-α)cos(2π-α)]/[cos(-π-α)tan(α+π)]
f(α)=(sinαcosα)/(-cosαtanα)
f(α)=(sinαcosα)/(-sinα)
f(α)=-cosα
f(-31π/3)=-cos(-31π/3)
=-cos(31π/3)
=-cos(2×5π+π/3)
=-cos(π/3)
=-1/2

高中数学三角函数化简题已知 fα=sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α)已知fα=sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α)化简若α第三象限,且cos(α-3/2π)=1/5,求fα若α=-31/3π,求f 已知f(α)=(tan(π-α)*cos(2π-α)*sin(二分之π+α))/(cos(-α-π)) (1)化简f(α) 已知f(α)=sin(π-α)cos(2π-α)tan(-α+3π/2)/cos(-π-α)则f(-13π/3)的值是 已知cosα+sinα/cosα+sinα 函数f(x)=3/2Xtan2α+3cos²α-sinαcosα 求函数f(x)解析式 1.已知tan(π+α)=3,则sin(π+α)×cos(π-α)= 2.已知sinα=根号5/5,则sinα的四次方-cosα的四次方=3..函数f(x)=sinπX/4,则f(1)+f(2)+...+f(2012)=4.已知sinα=1/3,则sinα的4次方-sinα的2次方+cosα的平方= cosα*cosβ=1,cos(α+β)等于sinαsinβ+cosαcosβ=0,则sinαcosα+sinβcosβ的值为已知tan(π/4+α)=2,求1/(2sinαcosα+(cosα)^2)的值求函数f(x)=2+2sinxcosx+sinx+cosx的最大值和最小值 sin(π/2+α)·cos(π/2-α)/cos(π+α)+sin(π-α)·cos(π/2+α)/sin(π+α)=?(2)sin²40°+sin²50°=?(3)已知sin10°=k,则cos620°=?(4)已知f(cosx)=cos3x,则f(cos30°)=? 已知f(sin+cos)=sin*cos,求f(sin30度)的值 若f(sinα+cosα)=sinαcosα,则f(cosπ/6)等于..麻烦讲清楚过程,辛苦各位了. 已知sinα+3cosα=2,求(sinα-cosα)/(sinα+cosα)的值 已知sinα+3cosα=2,求(sinα-cosα)/(sinα+cosα)的值 【三角函数】已知sin(α-π)=2cos(2π-α),求[sin(π-α)+5cos(2π-α)]/[3cos(π-α)-sin(-α)] 已知α∈(π,2π) sinα+cosα=1/5 求sinα*cosα sinα-cosα 已知α=7/12π,那么cosα√(1-sinα)/(1+sinα)+sinα√(1-cosα)/(1-cosα)= 已知tan(3π+α)=2,求:1、(sinα+cosα)²;2、sinα-cosα/2sinα+cosα 已知tan(π-α)=2,求sin²α-2sinαcosα-cos²α/4cos²α-3sin²α+1 已知tan(π-α)=2,求sinα-2sinαcosα-cosα/4cosα-3sinα的值 已知f(α)= sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α) (1)化简f(α) (2)若α...已知f(α)= sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α)(1)化简f(α)(2)若α为第二象限的角,且cos(α-3/2π)=1/5,求f(α)的值