不查表求值:cos5π/12·sinπ/12=1.如题2.根号下(cos4-sin^2(2)+2)=3.若sinα=4/5,tan(α+β)=1,α为第二象限角,则tanβ=4.sin2x/2cosx(1+tanx·tan(x/2))化简结果为
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sinπ/12cos5π/12等于
求cos5π/12sinπ/12
求值cos5π/8*cosπ8
为什么cos5π/12等于sinπ/12
cos5π/12=sinπ/12 怎么算的?
不查表求值:cos5π/12·sinπ/12=1.如题2.根号下(cos4-sin^2(2)+2)=3.若sinα=4/5,tan(α+β)=1,α为第二象限角,则tanβ=4.sin2x/2cosx(1+tanx·tan(x/2))化简结果为
sinπ/8*cos5π/8
2倍角公式求值》cosπ/12 x cos5π/12
求值:cos5π/8cosπ/8=
求值cos5π/8cosπ/8解cos5π/8cosπ/8,请写出过程.
求值cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11求详细过程和重要步骤的原因(1/4*sin8π/11*cos3π/11*cos5π/11)/2sinπ/11=(1/4*sin3π/11*cos3π/11*cos5π/11)/2sinπ/11这一步骤是怎么变过来的
(sin5π/12+cos5π/12)(sinπ/12-cosπ/12)
(sin5π/12-sinπ/12)(cos5π/12+cosπ/12)=?
cos5/12πcosπ/12+cosπ/12sinπ/6=?
sin(-4/3π)+cos5/6π=
cos5分之π×cos5分之2π
sin(-π/4)求值
计算不太明白...求值sin(-1320°)*cos1110°+cos(-1020°)sin750°+tan(-405°)求值sin25π/6+cos25π/3+tan25π/4+sin7π/3*cos13π/6-cos5π/3