(sinπ/12+cosπ/12)(sinπ/12-cosπ/12)=
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(sinπ/12+cosπ/12)(sinπ/12-cosπ/12)=(sinπ/12+cosπ/12)(sinπ/12-cosπ/12)=(sinπ/12+cosπ/12)(sinπ/12-cosπ
(sinπ/12+cosπ/12)(sinπ/12-cosπ/12)=
(sinπ/12+cosπ/12)(sinπ/12-cosπ/12)=
(sinπ/12+cosπ/12)(sinπ/12-cosπ/12)=
原式=sin²(π/12)-cos²(π/12)
=-cos(2×π/12)
=-cos(π/6)
=-√3/2
(sin5π/12-sinπ/12)(cos5π/12+cosπ/12)
=(cosπ/12 - sinπ/12)(sinπ/12 + cosπ/12)
=cos²(π/12) - sin²(π/12)
=cos(π/6)
=(根号3)/2
以上回答你满意么？
sin(α+π/3)+sinα=负5分之4根号3 α∈（-π/2,0）求cosα怎样从9/4*sin²α=9/4(1-cos²α)=3/4*cos²a+12/5*cosx+48/25化为cos²α+4/5*cosα-11/100=0呀sin(α+π/3)+sinα=-4√3/5sinαcosπ/3+cosαsinπ/3+sinα=-4√3/53/2*si
（sinπ/24)(cosπ/24)(cosπ/12)
（sinπ/24)(cosπ/24)(cosπ/12)
sinπ/12*cosπ/12=?
求值：sin(π/12)+cos(π/12)
sin(π/12)-cos(π/12)=
cos*π/12—sin*π/12
cosπ／12乘以sinπ／12
计算cosπ/12*sinπ/12
（cosπ/12-sinπ/12)*(COSπ/12+sinπ/12)等于
(cosπ/12 -sinπ/12)(cosπ/12 +sinπ/12)等于
（cosπ/12-sinπ/12)(cosπ/12+sinπ/12)的值为
（cosπ/12 -sinπ/12）（cosπ/12 +sinπ/12）=
(sinπ/12+cosπ/12)(sinπ/12-cosπ/12)=
已知α=7/12π,那么cosα√（1-sinα）/(1+sinα)+sinα√（1-cosα）/(1-cosα)=
cos^2π/12-sin^2π/12等于
sin(π/12)cos(π/12)的值
求sin π/12-(根号3)cos π/12