求数学高手帮忙cos(2π/7)+cos(4π/7))+cos(6π/7)=?(化简求值)
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cos(2π/7)+cos(4π/7)+cos(6π/7)
=[sin(2π/7)cos(2π/7)+sin(2π/7)cos(4π/7)+sin(2π/7)cos(6π/7)]/sin(2π/7)
(下面用积化和差）
=1/2[sin(4π/7)+sin(6π/7)-sin(2π/7)+sin(8π/7)-sin(4π/7)]/sin(2π/7)
=1/2[sin(6π/7)-sin(2π/7)+sin(8π/7)]/sin(2π/7)
=1/2[-sin(2π/7)]/sin(2π/7)
=-1/2
=[sin(2π/7)cos(2π/7)+sin(2π/7)cos(4π/7)+sin(2π/7)cos(6π/7)]/sin(2π/7)
(下面用积化和差）
=1/2[sin(4π/7)+sin(6π/7)-sin(2π/7)+sin(8π/7)-sin(4π/7)]/sin(2π/7)
=1/2[sin(6π/7)-sin(2π/7)+sin(8π/7)]/sin(2π/7)
=1/2[-sin(2π/7)]/sin(2π/7)
=-1/2
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