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求Cos20Cos40Cos80的值

cos20°cos40°cos80° =2sin20°cos20°cos40°cos80°/(2sin20°) =sin40°cos40°cos80°/(2sin20°) =sin80°cos80°/(4sin20°) =sin160°/(8sin20°) =sin20°/(8sin20°) =1/8;

不知道你是要求分开的值还是乘起来的值,如果是后者就简单了。 cos20°cos40°cos80° =2sin20°cos20°cos40°cos80°/2sin20° =sin40°cos40°cos80°/2sin20° =sin80°cos80°/4sin20° =sin160°/8sin20° =sin20°/8sin20° =1/8 谢谢望采纳

cos20°cos40°cos60°cos80° =(sin20cos20cos40cos60cos80)/sin20..........(乘以sin20°.再除以sin20°) =(sin40°cos40°cos60°cos80°)/2sin20°..........(2倍角公式) =(sin80°cos60°cos80°)/4sin20°................(2倍角公式) =(sin160°co...

cos20°cos40°cos80° =2sin20°cos20°cos40°cos80°/2sin20° =sin40°cos40°cos80°/2sin20° =sin80°cos80°/4sin20° =sin160°/8sin20° =sin20°/8sin20° =1/8

cos20°cos40°cos80° =2sin20°cos20°cos40°cos80°/(2sin20°) =sin40°cos40°cos80°/(2sin20°) =sin80°cos80°/(4sin20°) =sin160°/(8sin20°) =sin20°/(8sin20°) =1/8;

cos20°cos40°cos80°的值为______.-数学-魔方格 http://www.mofangge.com/html/qDetail/02/g0/201401/a90fg002450287.html 希望能帮助你

解: cos20*cos40*cos80 =(cos20*cos40)*cos80 =1/2(cos20-cos60)cos80 =1/2(cos20cos80-cos60cos80) =1/2[1/2(cos60-cos100)-1/2cos80] (因为cos100=-cos80) =1/4cos60=1/8 至于你的第二个问题,我想我比较有发言权,因为我是高一学生,而且...

十六分之一 cos20cos40cos60cos80 =2sin20cos20cos40cos80/4sin20 =sin40cos40cos60cos80/4sin20 =sin80cos80/8sin20 =sin160/16sin20 =1/16

cos20°*cos40°*cos80° =cos20°*cos40°*sin10° =(2cos10°sin10°cos20°cos40°)/2cos10° =(sin20°cos20°cos40°)/2cos10° =(sin40°cos40°)/4cos10° =sin80°/8cos10° =cos10°/8cos10° =1/8

cos20°cos40°cos60°cos80° =(sin20cos20cos40cos60cos80)/sin20.(乘以sin20°.再除以sin20°) =(sin40°cos40°cos60°cos80°)/2sin20°.(2倍角公式) =(sin80°cos60°cos80°)/4sin20°.(2倍角公式) =(sin160°cos60°)/8sin20°.(2倍角公式) =sin(1...

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