sin(α-β)cosα-coa(α-β)sinα=2/3 求cos2β和cos4βtan(2x+y)=3,tanx=-5 求tan(x+y)

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sin(α-β)cosα-coa(α-β)sinα=2/3 求cos2β和cos4βtan(2x+y)=3,tanx=-5 求tan(x+y)
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sin(α-β)cosα-coa(α-β)sinα=2/3 求cos2β和cos4βtan(2x+y)=3,tanx=-5 求tan(x+y)
sin(α-β)cosα-coa(α-β)sinα=2/3 求cos2β和cos4β
tan(2x+y)=3,tanx=-5 求tan(x+y)

sin(α-β)cosα-coa(α-β)sinα=2/3 求cos2β和cos4βtan(2x+y)=3,tanx=-5 求tan(x+y)
sin(α-β)cosα-coa(α-β)sinα=2/3
sin[(α-β)-α]=2/3
sin(α-β-α)=2/3
sin(-β)=2/3
-sinβ=2/3
sinβ=-2/3
cos2β
=1-2(sinβ)^2
=1-2*4/9
=1/9
cos4β
=2(cos2β)^2-1
=2*1/81-1
=-79/81
tan(2x+y)=3,tanx=-5 求tan(x+y)
tan(x+y)
=tan[(2x+y)-x]
=[tan(2x+y)-tanx]/[1+tan(2x+y)tanx]
=(3-5)/(1-3*5)
=-2/(-13)
=2/13

sin(α-β)cosα-coa(α-β)sinα=2/3
sin[(α-β)-α]=2/3
sinβ=2/3
cos2β=1-2sin²β=1-8/9=1/9
cos4β=2cos²2β -1=2*(1/81)-1=-79/81
tan(2x+y)=3,tanx=-5 求tan(x+y)
tan(x+y)=tan[(2x+y)-x]=[tan(2x+y)-tanx]/[1+tan(2x+y)tanx]=(3+5)/(1-3*5)=-4/7