神马作文网求值Cos pai/17 *cos 2pai/17*cos 4pai/17*cos 8pai/17
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求值Cos pai/17 *cos 2pai/17*cos 4pai/17*cos 8pai/17
cos(π/17)cos(2π/17)cos(4π/17)cos(8π/17)
=[16sin(π/17)cos(π/17)cos(2π/17)cos(4π/17)cos(8π/17)]/16sin(π/17)
=[8sin(2π/17)cos(2π/17)cos(4π/17)cos(8π/17)]/16sin(π/17)
=[4sin(4π/17)cos(4π/17)cos(8π/17)]/16sin(π/17)
=[2sin(8π/17)cos(8π/17)]/16sin(π/17)
=sin(16π/17)/16sin(π/17)
=1/16
=[16sin(π/17)cos(π/17)cos(2π/17)cos(4π/17)cos(8π/17)]/16sin(π/17)
=[8sin(2π/17)cos(2π/17)cos(4π/17)cos(8π/17)]/16sin(π/17)
=[4sin(4π/17)cos(4π/17)cos(8π/17)]/16sin(π/17)
=[2sin(8π/17)cos(8π/17)]/16sin(π/17)
=sin(16π/17)/16sin(π/17)
=1/16
cos(pai/15)×cos(2pai/15)×cos(3pai/15)×cos(4pai/15)×cos(5pai/
f x =sin(pai*x/4-pai/6)-2(cos pai*x/8)^2+1
sin(pai-a)cos(-8pai-a)=60/169且a属于(pai/4,pai/2)求cosa,sina
[tan(pai-a)cos(2pai-a)sin(-a+3pai/2)]/[cos(-a-pai)sin(-pai-a
化简：[cos(a+pai)*sin^2(a+pai)]/[tan^2(pai+a)*cos^3a]
化简cos(pai/4 +a)+sin(pai/4 +a)
f(a)=[sin(pai-a)cos(2pai-a)tan(-a+3pai/2)]/[cos(-pai-a)]则f(-
设tan（pai+a）=2,则sin（a-pai）+cos（pai-a）/sin（pai+a）-cos（pai+a）等于
化简!f(x)=sin(pai-x)cos(3/2pai+x)+sin(pai+x)sin(3/2pai-x)
已知：cos(x-pai/4)=根号2/10,x属于pai/2,3pai/4,求sinx值
已知a属于(pai/2,pai),tan(a+pai/4)=1/7,sin a+cos a=?
化简sin(-2/3pai+kpai)cos(pai/6+kpai)tan(pai/4+kpai),k属于z