cos(pai/15)×cos(2pai/15)×cos(3pai/15)×cos(4pai/15)×cos(5pai/
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/10 11:03:13
cos(pai/15)×cos(2pai/15)×cos(3pai/15)×cos(4pai/15)×cos(5pai/15)×cos(6pai/15)×cos(7pai/15)求值
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
=[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
cos(pai/15)×cos(2pai/15)×cos(3pai/15)×cos(4pai/15)×cos(5pai/
[tan(pai-a)cos(2pai-a)sin(-a+3pai/2)]/[cos(-a-pai)sin(-pai-a
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f(a)=[sin(pai-a)cos(2pai-a)tan(-a+3pai/2)]/[cos(-pai-a)]则f(-
f x =sin(pai*x/4-pai/6)-2(cos pai*x/8)^2+1
化简cos(pai/4 +a)+sin(pai/4 +a)
已知cos(3pai/2-a)=2根号5/5,a属于(pai,3pai/2) 求tan(pai/4+2a)的值
化简!f(x)=sin(pai-x)cos(3/2pai+x)+sin(pai+x)sin(3/2pai-x)
sin(pai-a)cos(-8pai-a)=60/169且a属于(pai/4,pai/2)求cosa,sina
cos(a+b)=3/5 sin(b-pai/4)=5/13 a,b 属于(0,pai/2) 求cos(a+pai/4)
cos(pai/3+a)=1/3,cos(2pai/3+2a)=?
若cos(pai/3+a)=-1/2,则cos(2pai/3+2a)=?