高一数学(两倍角)求值：COSa
估计题目中的a是π。
1)预备:sin2A=2sinAcosA--->cosA=sin2A/sinA
cos(π/17)cos(2π/17)cos(4π/17)cos(8π/17)
=sin(2π/17)/[2sin(π/17)]*sin(4π/17)/[2sin(2π/17)]
*sin(8π/17)/[2sin(4π/17)]*sin(16π/17)/[2sin(8π/17)]
=sin(16π/17)/[2^4*sin(π/17)]
=sin(π/17)/[16sin(π/17)
=1/16
2)cos(11π/12)*sin(5π/12)
=-cos(π/12)*cos(π/1...全部
估计题目中的a是π。
1)预备:sin2A=2sinAcosA--->cosA=sin2A/sinA
cos(π/17)cos(2π/17)cos(4π/17)cos(8π/17)
=sin(2π/17)/[2sin(π/17)]*sin(4π/17)/[2sin(2π/17)]
*sin(8π/17)/[2sin(4π/17)]*sin(16π/17)/[2sin(8π/17)]
=sin(16π/17)/[2^4*sin(π/17)]
=sin(π/17)/[16sin(π/17)
=1/16
2)cos(11π/12)*sin(5π/12)
=-cos(π/12)*cos(π/12)
=-[cos(π/12)]^2
=-[1+cos(π/6)]/2
=-(1+√3/2)/2
=-(2+√3)/4
3)tgA+tgB+3=3tgAtgB--->tgA+tgB=-3+3tgAtgB
--->tgA+tgB=-3(1-tgAtgB)
--->(tgA+tgB)/(1-tgAtgB)=-3
--->tg(A+B)=-3
A+B=π-C--->tg(A+B)=-tgC
----tgC=-tg(A+B)=-(-3)=3。收起
