cos(pai/15)×cos(2pai/15)×cos(3pai/15)×cos(4pai/15)×cos(5pai/

junny8882022-10-04 11:39:542条回答

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

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linaww 共回答了24个问题 | 采纳率91.7%: 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
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