3 14-3 7cos²5π 12

原式=根号2*(根号2/2*cosπ/12—根号2/2*sinπ/12)=根号2*(sinπ/4cosπ/12—cosπ/4sinπ/12)=根号2*sin(π/4-π/12)=根号2*sin(π/6

＝（1/2）sinπ/6＝（1/2）×（1/2）＝1/4这题是代入sin2α＝2sinαcosα求得的

/>cos(5π/6+a)=cos[π-(π/6-a)]=-cos(π/6-a)=-√3/3cos²(π/3+a)=cos²[(π/2)-(π/6-a)]=sin²(π/

原式=[2sinxcosx＋2sin²x]/[(cosx－sinx)/cosx]=[2sinxcosx(sinx＋cosx)]/[cosx－sinx]=[sin2x][(sinx＋cosx)

[sin(π/12)+cos(π/12)]^2=[sin(π/12)]^2+2sin(π/12)*cos(π/12)+[cos(π/12)]^2=1+22sin(π/12)*cos(π/12)=1+s

cos(π/12)*cos(5π/12)=cos(π/12）*cos(π/2-π/12)=cos(π/12)*sin(π/12)=(1/2)*sin(π/6)=(1/2)*(1/2)=1/4

证明cos5π/12=cos(π/2-5π/12)=sinπ/12∴cos²5π/12+cos²π/12=sin²π/12+cos²π/12=1（cos5π/1

cos²5π/12+cos²π/12+cosπ/12*cos5π/12由于cos5π/12=sin（π/2-5π/12）=sinπ/12原式=sin²π/12+cos&#

cos^2(π/4+π/6)+cos^2(π/4-π/6)+1/2[cos(5π/12+π/12)+cos(5π/12-π/12)]=cos^2(π/4+π/6)+cos^2(π/4-π/6)+1/2

cos(a+5π/12)cos(a+π/6)+cos(π/12-a)cos(π/3-a）=sin(π/2-(a+5π/12))cos(a+π/6)+cos(π/12-a)sin(π/2-(π/3-a）

因为cos(α)=cos((2α+β)-(α+β))=cos(2α+β)cos(α+β)-sin(2α+β)sin(α+β),又cos(α+β)=12/13,cos(2α+β)=3/5且β∈(0,π/

解本题就是证明cos^2（75°）+cos^2（15°）+cos（75°）cos（15°）=5/4我们用分析法证明欲证cos^2（75°）+cos^2（15°）+cos（75°）cos（15°）=5/

因为cos(π+A)＝−12，所以cosA=12，sin(π2+A)=cosA=12．故答案为：12．

利用黄金分割容易得COSπ/5=(根号5+1)/4COS2π/5=(根号5-1)/4乘起来是1/4

cos(π/12-α)-√3cos(5π/12+α)=2[(1/2)cos(π/12-α)-(√3/2)sin(π/12-α)](∵cos(5π/12+α)=sin[π/2-(5π/12+α)]=si

原式=2sin(π/10)cos(π/10)cos(π/5)/2cos(π/10)=sin(π/5)cos(π/5)/2cos(π/10)=2sin(π/5)cos(π/5)/4cos(π/10)=s

f(x)=cos^2x/sin(π/4+x)cos(x+π/4),f(5π/12)=cos²(5π/6)/[(sin2π/3)cos(2π/3)]=(3/4)/(-√3/2*1/2)=-√3

∵5π12+θ+π12-θ=π2，∴π12-θ=π2-（5π12+θ），∴sin（π12-θ）=sin[π2-（5π12+θ]=cos（5π12+θ），∴sin（π12-θ）=cos（5π12+θ）=

∵−π＜α＜−π2∴−7π12＜α+ 5π12＜−π12∵cos(5π12+α)＝13∴sin(α+5π12)＝−223∵(5π12+α)+(π12−α)＝π2，∴cos(π12−α)＝co

原式=2sinπ/5cosπ/5cosπ/5/(2sinπ/5)=sin2π/5cosπ/5/(2sinπ/5)=2sin2π/5cosπ/5/(4sinπ/5)=sin4π/5/(4sinπ/5)=