Question

高一数学题

已知：2tana=3tanb求证tan(a-b)=sin2b/(5-coa2b),

Answer 1

2tana=3tanb => 2sina/cosa=3sinb/cosb => sina*cosb=3sin(a-b),cosa*sinb=2sin(a-b);
欲证乎枯tan(a-b)=(sin2b)/(5-cosb)
只需(sina*cosb-sinb*cosa)/(cosa*cosb+sina*sinb)=2sinb*cosb/[5-(cosb)^2+(sinb)^2]
只需枣顷并sin(a-b)/(cosa*cosb+sina*sinb)=2sinb*cosb/[5-(cosb)^2+(sinb)^2]
只需2cosa*sinb*(cosb)^2+2sina*cosb*(sinb)^2=sin(a-b)*[5-(cosb)^2+(sinb)^2]
只需4sin(a-b)*(cosb)^2+6sin(a-b)*(sinb)^2=5sin(a-b)-sin(a-b)*(cosb)^2+sin(a-b)*(sinb)^2
只需5sin(a-b)*(cosb)^2+5sin(a-b)*(sinb)^2=5sin(a-b)
因为(cosb)^2+(sinb)^2=1
所以可凳迹证：若2tana=3tanb,则tan(a-b)=(sin2b)/(5-cos2b)

Answer 2

用万能公带仿式吧:
sin2θ=2tanθ/[1+(tanθ)^2]
cos2θ=[1-(tanθ)^2]/[1+(tanθ)^2]
tan2θ=2tanθ/[1-(tanθ)^2]

所以sin2b/(5-cos2b)={2tana/[1+(tana)^2]}/{5-[1-(tana)^2]/[1+(tana)^2]}=tana/[2+3(tana)^2]......(1)
而tan(a-b)=(tana-tanb)/(1-tana*tanb)
把tanb=(2*tana)/3代尘圆入(tana-tanb)/(1-tana*tanb)即可发现等派行塌于(1)式.

Answer 3

高中数学
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