cos(-19/6π)=?

cos(-19/6π)=? 化简cos(π/6a)+cos(π／6-a)= cos(π/7) + cos(6π/7) =几 cos(- 19/4 π)=? cos(-19／6n)= cos 2π/7 +cos 4π/7 +cos 6π/7=? 请问cos(2π/7)+cos(4π/7)+cos(6π/7)=?, cos(π/32)*cos(π/16)*cos(π/8)= cos(19π/6)+tan(16π/3)=? 由cos(a+b)=cos a cos b-sin a sin b cos(a-b)=cos a cos b+sin a sin b解题设a为锐角,证：1、2分之根3乘cos a + 2分之1乘sin a=cos(6分之π-a)2、cos a-sin a=根号2cos(4分之π+a) cos(2π/7)+cos(4π/7)+cos(6π/7)= -1/2.证明[cos(2π/7)]^2+[cos(4π/7)]^2+[cos(6π/7)]^2=5/4 cos(2π/7)+cos(4π/7)+cos(6π/7)= -1/2.证明[cos(2π/7)]^2+[cos(4π/7)]^2+[cos(6π/7)]^2=5/4 求数学高手帮忙cos(2π/7)+cos(4π/7))+cos(6π/7)=?(化简求值) cosπ/6=(),sinπ=() 已知cos(x-π/6)=-根号三分之三 则cos(x)+cos(x-π/3)的值是 cos（-19/6∏）= cos(-17/6π）= cosπ/12=