cos(pai/15)×cos(2pai/15)×cos(3pai/15)×cos(4pai/15)×cos(5pai/15)×c_数学解答

cos(pai/15)×cos(2pai/15)×cos(3pai/15)×cos(4pai/15)×cos(5pai/15)×cos(6pai/15)×cos(7pai/15)求值

人气：378 ℃　时间：2020-07-09 15:00:39

解答

原式=[1/(sinπ/15)]•[sin(π/15)cos(π/15)cos(2π/15)．．cos(7π/15)]=[sin(8π/15)/8sin(π/15)]•[cos(3π/15)cos(5π/15)cos(6π/15)cos(7π/15)]=．．．=[sin(4π/5)/2^7sin(π/5)]=1/128字数限制...