sin3/7π*cos2/5π-cos(-3/7π)*sin(-2/5π)

sin3/7π*cos2/5π-cos(-3/7π)*sin(-2/5π) (sinα+sin2α+sin3α)/(cosα+cos2α+cos3α)= - 4cotα且π/2 cos2π/3(sin3π/5+icos3π/5)的三角式是 求COSπ/5*COS2π/5 cosπ/5×cos2π/5= COSπ/5乘以COS2π/5 cosπ/5×cos2π/5 cosπ/5cos2π/5 cosπ/5cos2π/5=? cosπ/5 - cos2π/5 怎样变成 cosπ/5 + cos2π/5rt 解三角方程,cosβ + cos2β = sin3β 求值:cosπ/7*cos2π/7*cos3π/7 cosπ/7cos2π/7cos3π/7=? cosπ/7cos2/7πcos4/7π= sin3/5=cos?详解. 已知(sinα+sin2α+sin3α)/(cosα+cos2α+cos3α)=﹣4/tanα,且π＜2α＜3π/2,求cosα的值. （cosπ/5）（cos2π/5)的值等于 cosπ/5·cos2π/5等于多少?