cos20cos40cos80=?求过程!
解:由和差化积公式,
cos 40° +cos 80° =2 cos 60° cos 20°
=cos 20°.
所以 原式 =cos 20° -(cos 40° +cos 80°)
=0.
= = = = = = = = =
百度百科:和差化积公式。
不记得的话,用和角公式和差角公式自己推。
cos20°cos40°cos80°
=2sin20°cos20°cos40°cos80°/2sin20°
=sin40°cos40°cos80°/2sin20°
=sin80°cos80°/4sin20°
=sin160°/8sin20°
=sin20°/8sin20°
=1/8
=sin20cos20cos40cos80/sin20
=1/2*sin40cos40cos80/sin20
=1/4*sin80cos80/sin20
=1/8*sin160/sin20
=1/8*sin20/sin20
=1/8
不懂的欢迎追问,如有帮助请采纳,谢谢!
用倍角公式的逆用
cos20°°cos40°os80°的结果等于多少?
cos20°cos40°cos80°的结果等于1\/8。解:cos20°cos40°cos80° =cos20°cos40°cos80°*sin80°\/sin80° =1\/2*cos20°cos40°*sin160°\/sin80° =1\/2*cos20°cos40°*sin(180°-160°)\/sin80° =1\/2*sin20°cos20°cos40°\/sin80° =1\/4*sin40°cos40°\/sin80° =...
已知锐角α的终边上一点P(sin40°,1+cos40°)则锐角α=( ) A.80°...
由题意可知sin40°>0,1+cos40°>0,点P在第一象限,OP的斜率tanα= 1+cos40° sin40° = 1+2c os 2 20°-1 2sin20°cos20° =cot20°=tan70°,由α为锐角,可知α为70°.故选B.
中学平面几何难题,求解! http:\/\/wenku.baidu.com\/view\/fb05bd86d4d8...
DE=a*sin(20)=2*BC*sin(20)*sin(20)z=CB*cos(40)x\/z=(sin(20))^2\/[cos(30)*cos(40)]cos(30)*cos(40)=1\/2[(cos(70)+cos(10)]=1\/2[(sin(20)+sin(80)]=1\/2sin(20)[1+4cos(20)cos(40)]=1\/2sin(20)[1+4(1\/2(cos(60)+cos(20)))]=1\/2sin(20)[1+4(1...
爰古康利:[答案] cos20cos40cos80 =sin20°cos20°cos40°cos80°/sin20° =(1/2)sin40°cos40°cos80°/sin20° =(1/4)sin80°cos80°/sin20° =(1/8)sin160°/sin20° =(1/8)sin20°/sin20° =1/8
拜泉县17738997279: cos20cos40cos80=?化简求值 - ?
爰古康利: 解:cos20cos40cos80 =sin20cos20cos40cos80/sin20 =1/2*sin40cos40cos80/sin20 =1/4*sin80cos80/sin20 =1/8*sin160/sin20 =1/8*sin20/sin20 =1/8
拜泉县17738997279: cos20cos40cos80 写出具体过程, - ?
爰古康利:[答案] 原式=sin20cos20cos40cos80/sin20=sin160/8sin20=1/8
拜泉县17738997279: 20度,40度,80度的余弦值相乘?答对赞. - ?
爰古康利:[答案] COS20COS40COS80=SIN20COS20COS40COS80/SIN20=1/2SIN40COS40COS80/SIN20 =1/4SIN80COS80/SIN20=1/8SIN160/SIN20=1/8SIN20/SIN20=1/8
拜泉县17738997279: cos20cos40cos80=what - ?
爰古康利: =sin20cos20cos40cos80/sin20=0.5sin40cos40cos80/sin20=0.25sin80cos80/sin20=0.125sin160/sin20=0.125
拜泉县17738997279: cos20cos40cos60cos80等于多少 - ?
爰古康利:[答案] 十六分之一 cos20cos40cos60cos80 =2sin20cos20cos40cos80/4sin20 =sin40cos40cos60cos80/4sin20 =sin80cos80/8sin20 =sin160/16sin20 =1/16
拜泉县17738997279: cos20cos40cos80的值?单位是度 - ?
爰古康利:[答案] cos20°cos40°cos80° =2sin20°cos20°cos40°cos80°/2sin20° =sin40°cos40°cos80°/2sin20° =sin80°cos80°/4sin20° =sin160°/8sin20° =sin20°/8sin20° =1/8
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爰古康利:[答案] cos20cos40cos80 =sin20cos20cos40cos80/sin20 =sin40cos40cos80/(2sin20) =sin80cos80/(4sin20) =sin160/(8sin20) =sin(180-160)/(8sin20) =sin20/8/sin20 =1/8
拜泉县17738997279: cos20°cos40°cos80°=多少,求完整解题,还有用到哪个公式 - ?
爰古康利:[答案] cos20cos40cos80 =sin20cos20cos40cos80/sin20 =(1/8)sin160/sin20=(1/8)
拜泉县17738997279: cos20cos40cos80怎么算,要过程,那个20 40 80是度数同上 - ?
爰古康利:[答案] sin20分之cos20sin20cos40cos80.然后用公式2sinacosa=sin2a就可以了