作业帮cos20+cos60+cos100+cos140求值
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cos20+cos60+cos100+cos140求值
cos20+cos60+cos100+cos140
求值
cos20+cos60+cos100+cos140求值
原式=cos20+cos60+cos(120-20)+cos(120+20)
=cos20+cos60+cos120cos20+sin120sin20+cos120cos20-sin120sin20
=cos20+cos60+2cos120cos20
=cos20+cos60-cos20
=cos60
=1/2
cos20+cos60+cos100+cos140求值
cos20度+cos60度+cos100度+cos140度=?
数学cos20+cos60+cos100+cos140=?如题过程
cos20°-sin40°+cos60°+cos100°
cos20°—cos40°+cos60°+cos100°的值答案为1/2
设M＝cos20°×cos40°×cos60°×cos80°×cos100°×cos120°×cos140°×cos160°,求证：M＝1/256
已知logaX=1/cos20°,logbX=1/cos60°,logcX=1/cos100°,logdX=1/cos140°,求logabcdX的值
cos20°+cos100度+cos140度=?
求cos20×cos40×cos60×cos80
sin10 *cos20*cos40*cos60 的值
化简求值：(sin60/sin20)-(cos60/cos20)
cos20度+cos100度-cos40度等于多少求过程
(cos20)∧4+(cos100)∧4+(cos140)∧4
化简cos20°cos40°cos60°cos80°
cos20°*cos40°*cos60°*cos80°=
求值：cos20度cos40度cos60度cos80度＝?
COS20度*COS40*COS60*COS80=?请写过程
cos20°cos40°cos60°cos80°是多少