已知f(α)=sin(π-α)cos(2π-α)tan(-α 3π 2)

sin(π-a)=sinacos(2π-a)=cosatan(1.5π-a)=cotatan(-a-π)=-tanasin(-π-a)=sina所以f(a)=sinacosacota(-tana)/s

f(α)=sin(π-α)cos(2π-α)tan(-α-π)/tan（-α）sin（-π-α）=sinα*cosα*(-tanα)/[(-tanα)*sinα]=cosαcos(α-3/2π)=-c

f(a)=sin(π-α)cos(2π-α)tan(-α+π)/sin(π+α)=sinacosa(-tana)/(-sina)=sinacosasina/cosa*1/sina=sinacos(a-

∵fα=sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin（-π-α）∴f(a)=-sinacosasina/(sinasina)=-cosa∵cos(a-3/2π

sinα=1/2,α为锐角∴cosα=√(1-sin²α)=√(1-1/4)=√3/2∴f(α)=sin(α+π/6)=sinα*cos(π/6)+cosα*sin(π/6)=(1/2)*(

1.f(α)=[sin(π-α)cos(2π-α)]/sin(3π-α)=sinacosa/sina=cosa2.α是第三象限角,sinα=-3/5,cosa=-4/5,f（α）=-4/53.f(α)

f(α)=sin(π-α)cos(2π-α)tan(-α+π)/sin(π+α)tan(2π-α)=sinacosa(-tana)/(-sina)(-tana)=-(sinacosatana/sina

（1）f(α)=[sin(π+α)cos(2π-α)tan(-α+3π/2)tan(-α-π)]/sin(π-α)=[(-sinα)*cosα*cotα*(-tanα)]/sinα=cosα；（2）∵

⑴f(α)=[sin(π-α)cos(2π-α)tan(-π-α)]/sin(-π-α)=[sin(α)cos(-α)tan(-α)]/sin(α)=cos(-α)tan(-α)=cos(-α)*si

(1)原式=sin²a*cosa*tana/(-sina)*(-tana)=sina*cosa(2)f(a)=sina*cosa=1/8∵π/4

f(a)=-cosasina=-1/5,cosa=-2根号6/5,f(a)=2根号6/5直接代入f(a)=f(-7派/6）=-根号3/2

（根号2＋根号6）÷4再问：如何做的？？？？？？？？？？谢谢

(1)cos2α=cos2α/1=cos2α/[(cosα)^2+(sinα)^2]因为cos2α=(cosα)^2-(sinα)^2所以原式=[=(cosα)^2-(sinα)^2]/[(cosα)

f(α)=(sinα*cosα*-tanα)/tan(π+α)*sin(π-α)=(-sinα*cosα*tanα)/tanα*sinα=-cosαcos(α-3π/2)=cos(α-3π/2+2π)

f(α)=(sinα*cosα*-tanα)/tan(π+α)*sin(π-α)=(-sinα*cosα*tanα)/tanα*sinα=-cosαcos(α-3π/2)=cos(α-3π/2+2π)

f(x)=sin(πx/2+α),f(2009)=sin(2009π/2+a)=1所以a=2kπf(2010)=sin(1005π+2kπ)=sin(π)=0再问：所以a=2kπf(2010)=sin

凑出来了,好题目思想如下：用n项的均值不等式,然后再用辅助角公式,等号成立条件有两个,但是n也是参数,所以等于是两个方程两个未知数,可能有解,且听我细细道来首先8/cosα份数挺多,我想把它拆分成n等

1,f(α)=sin(π/2+α)+3sin(-π-α)/2cos(11π/2-α)-cos(5π-α).=(cosα-3sinα)/(-2sinα+cosα)f(α)=(cosα-3sinα)/(c

解由题知函数f(x)=sinx在x∈[-π\2,π\2]是奇函数且单调递增则由f(sinα)+f(cosα-1\2)=0得f(sinα)=-f(cosα-1\2)即f(sinα)=f(-cosα+1\

(1)f(a)=sina*cosa*cota*(-tana)/(sina)=-cosa(2)cos(a-3π/2)=1/5即-sina=1/5∴sina=-1/5∴cos=-2√6/5即f(a)=2√