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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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
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

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
1 ,cos5π/12·sinπ/12
=cos(π/4+π/6)*sin(π/4-π/6)
=(√6-√2)/4*(√6+√2)/4=1/4
2.√[cos4-sin^2(2)+2]
=√[cos^2(2)-2sin^2(2)+2]
=√[cos^2(2)+2(1-sin^2(2)]
=√[3cos^2(2)]=-√3cos2
3;sinα=4/5,cosα=-3/5 ,tanα=-4/3
tan(α+β)=1=(tanα+tanβ)/(1-tanαtanβ)
(1-tanαtanβ)=(tanα+tanβ)
1+4/3tanβ=-4/3tanβ ,tanβ=-3/8
4

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