已知cos(x-六分之π)=-三分之根号三,则cosx+cos(x三分之π)=？
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cos(x-π/6)=-√3/3
cosx + cos(x-π/3) = cos(x-π/6 + π/6 ) + cos(x-π/6-π/6)
= cos(x-π/6)cosπ/6 - sin(x-π/6)sinπ/6 + cos (x-π/6)cosπ/6 + sin(x-π/6)sinπ/6
=2cos(x-π/6)cosπ/6
=2*(-√3/3 ) * ( √3/2 )
=-1,8,
cosx + cos(x-π/3) = cos(x-π/6 + π/6 ) + cos(x-π/6-π/6)
= cos(x-π/6)cosπ/6 - sin(x-π/6)sinπ/6 + cos (x-π/6)cosπ/6 + sin(x-π/6)sinπ/6
=2cos(x-π/6)cosπ/6
=2*(-√3/3 ) * ( √3/2 )
=-1,8,
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