问题: 求对数的值
㏒2 cosπ/9+ ㏒2 cos2π/9+ ㏒2 cos4π/9 =
( 对数的底为2 )
解答:
㏒2 cosπ/9+ ㏒2 cos2π/9+ ㏒2 cos4π/9
=㏒2(cosπ/9)(cos2π/9)(cos4π/9)
=㏒2[(sinπ/9)(cosπ/9)(cos2π/9)(cos4π/9)/(sinπ/9)]
=㏒2[(sin2π/9)(cos2π/9)(cos4π/9)/2(sinπ/9)]
=㏒2[(sin4π/9)(cos4π/9)/4(sinπ/9)]
=㏒2[(sin8π/9)/8(sinπ/9)]
=㏒2(1/8)
=-3
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