已知tan(x 5π 4)=2
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由已知有tanα+tanβ=-33tanα•tanβ=4,…(2分)∴tan(α+β)=tanα+tanβ1-tanαtanβ=-331-4=3,…(5分)∵tanα•tanβ=4>0,tanα
tan(a-π/4)=(tana-tanπ/4)/(1+tana*tanπ/4)=(2-1)/(1+2*1)=2/3
sn=(2-3x5^-1)+(4-3x5^-2)+…+(2n-3x5^-n)=(2+4+.+2n)-(3x5^-1+3x5^-2+.+3x5^-n)=(2^1+2^2+.+2^n)-(3x5^-1+3
tan(a+π/4)=(tana+tanπ/4)/1-tana*tanπ/4=(2+1)/1-2*1=-3
(1)∵tanα=2,∴tan(α+π4)=tanα+11−tanα=2+11−2=-3.(2)∵tanα=2,∴6sinα+cosα3sinα−2cosα=6tanα+13tanα−2=12+16−
/>(1)tan(α+β-π/4)=tan(π/12+α+β-π/3)=[tan(π/12+α)+tan(β-π/3)]/[1-tan(π/12+α)tan(β-π/3)]=(√2+2√2)/[1-√
1)π/4
1.tan(π/4+α)=2[tan(π/4)+tanα]/[1-tan(π/4)*tanα]=(1+tanα)/(1-tanα)=2tanα=1/32.(sin2α·cosα-sinα)/(sin2
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1.∵tan(a/2)=2∴tana=[2tan(a/2)]/{1-[tan(a/2)]^2}=(2×2)/(1-2^2)=-4/3∴tan(a+π/4)=[tana+tan(π/4)]/[1-tan
1.tan(a+π/4)=tan[(a+p)-(p-π/4)]=[tan(a+p)-tan(p-π/4)]/[1+tan(a+p)*tan(p-π/4)]=[2/5-1/4]/[1-2/5*1/4]=
左边=(tana-tanb)/(-1/tana+cotb)=(tana-tanb)/(-1/tana+1/tanb)上下乘tanatanb=tanatanb(tana-tanb)/(tana-tanb
∵tan(α+β)=4tanβ,∴tanα+tanβ1−tanαtanβ=4tanβ,∴4tanαtan2β-3tanβ+tanα=0,①∴α,β∈(0,π2),∴方程①有两正根,tanα>0,∴△=
tan(a+π/4)=(tana+tan(π/4))/[1-tana*tan(π/4)]=(3+1)/(1-3*1)=-2tan(a-π/4)=(tana-tan(π/4))/[1+tana*tan(
tanθ:tan(π/4+θ)=tanθ:(1+tanθ)/(1-tanθ)=2:15所以tanθ=1/5或tanθ=2/3则对应的tan(π/4+θ)=3/2或tan(π/4+θ)=5所以p=-(t
2tanθ/1+tan^2θ=3/5即tan2θ=3/5sin2θ=3/根号(34)cos2θ=5/根号(34)sin^2(π/4+θ)=(1-cos(2θ+π/2))/2=(1+sin(2θ))/2
tan(π/4+a)=(tanπ/4+tana)/(1-tanπ/4*tana)=(tana+1)/(1-tana)=3tana+1=3-3tanatana=1/2sin2a-2cos^2a-1=si
tan(π/4+a)=(1+tana)/(1-tana)=-1/2tana=-3sina=3根号10/10[sin2a-2(cosa)^2]/1+tana=[sin2a-2+2(sina)^2]/1+
1*5+2*5+...+n*5=(1+2+...+n)*5=n(n+1)*5/2