√3cosπ/12-sinπ/12=
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√3cosπ/12-sinπ/12=
√3cosπ/12-sinπ/12=
√3cosπ/12-sinπ/12=
√3cosπ/12-sinπ/12
=2*[cos(π/12)*(√3/2)-sin(π/12)*(1/2)]
=2*[cos(π/12)*cos(π/6)-sin(π/12)*(sin(π/6)]
=2*cos(π/12 +π/6) (两角和的余弦公式)
=2*cos(π/4)
=√2
√3cosπ/12-sinπ/12=
若方程12x²+πx-12x=0的两的根分别是α,β,则cosαcosβ-√3sinαcosβ-√cosαsinβ-sinαsinβ=
sinπ/12*cosπ/12=?
sin(π/12)-cos(π/12)=
已知α=7/12π,那么cosα√(1-sinα)/(1+sinα)+sinα√(1-cosα)/(1-cosα)=
sinπ/12-√3cosπ/12的值
根号3×cosπ/12-sinπ/12=?
化简sinπ/12-√3cosπ/12化简sinπ/12-√3cosπ/12
(cosπ/12 -sinπ/12)(cosπ/12 +sinπ/12)=
(sinπ/12+cosπ/12)(sinπ/12-cosπ/12)=
tanα=3,则sinα-cosα/2sinα+cosα等于多少-π/12等于多少
1.已知cosα=12/13,α∈(3π/2,2π),求cos(α+π/4).2.cos(α+β)=1/3,则(sinα-sinβ)^2+(cosα+cosβ)^2=?3.已知sinα=(√5)/5,sinβ=(√10)/10,α、β都是锐角,求cos(α+β)及α+β的值.4.已知锐角α、β满足sinα=(√5)/5,cosβ(√10)/1
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
化简 cos (π/12)+√3sin(π/12)=
sin(5π/12)-√3cos(5π/12)=
(sinπ/24)(cosπ/24)(cosπ/12)
(sinπ/24)(cosπ/24)(cosπ/12)
请会的大哥大姐多多指教.已知sinθ=(√5-1)/4(1)求(sinθ-cosθ)/(sinθ+cosθ)+(sinθ+cosθ)/(sinθ-cosθ)的值.(2)已知5sinθ+12cosθ=0,求(sinθ+9cosθ)/2(2-3sinθ)的值.