sin3/7π*cos2/5π-cos(-3/7π)*sin(-2/5π)
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sin3/7π*cos2/5π-cos(-3/7π)*sin(-2/5π)
sin3/7π*cos2/5π-cos(-3/7π)*sin(-2/5π)
sin3/7π*cos2/5π-cos(-3/7π)*sin(-2/5π)
sin3/7π*cos2/5π-cos(-3/7π)*sin(-2/5π)
=sin3/7π*cos2/5π+cos(3/7π)*sin(2/5π)
=sin(3/7π+2/5π)
=sin(49π/35)
=-sin(14π/35)
=-sin(2π/5)
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?