已知sinB=msin(2a+B)且m≠1,a≠k丌/2,a﹢B≠丌/2+k丌﹙k∈Z﹚求证:tan﹙a+B﹚=﹙1﹢m
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已知sinB=msin(2a+B)且m≠1,a≠k丌/2,a﹢B≠丌/2+k丌﹙k∈Z﹚求证:tan﹙a+B﹚=﹙1﹢m/l﹣m﹚tana
你想求证的结论是 tan(a+B)=[(1+m)/(1-m)]tana 吧!若是这样,则方法如下:
∵sinB=msin(2a+B),∴m=sinB/sin(2a+B),
∴1+m=[sinB+sin(2a+B)]/sin(2a+B),······①
1-m=[sin(2a+B)-sinB]/sin(2a+B),······②
①÷②,得:(1+m)/(1-m)=[sinB+sin(2a+B)]/[sin(2a+B)-sinB].······③
而tan(a+B)/tana
=sin(a+B)cosa/[cos(a+B)sina]
=(sinacosB+cosasinB)cosa/[(cosacosB-sinasinB)sina]
=[sinacosacosB+(cosa)^2sinB]/[sinacosacosB-(sina)^2sinB]
=[2sinacosacosB+2(cosa)^2sinB]/[2sinacosacosB-2(sina)^2sinB]
=[sin2acosB+(1+cos2a)sinB]/[sin2acosB+(-1+cos2a)sinB]
=(sin2acosB+cos2asinB+sinB)/(sin2acosB+cos2asinB-sinB)
=[sin(2a+B)+sinB]/[sin(2a+B)-sinB].······④
比较③、④,得:(1+m)/(1-m)=tan(a+B)/tana
∴tan(a+B)=[(1+m)/(1-m)]tana.
注:若需要证明的结论不是我所猜测的那样,则请你补充说明.
∵sinB=msin(2a+B),∴m=sinB/sin(2a+B),
∴1+m=[sinB+sin(2a+B)]/sin(2a+B),······①
1-m=[sin(2a+B)-sinB]/sin(2a+B),······②
①÷②,得:(1+m)/(1-m)=[sinB+sin(2a+B)]/[sin(2a+B)-sinB].······③
而tan(a+B)/tana
=sin(a+B)cosa/[cos(a+B)sina]
=(sinacosB+cosasinB)cosa/[(cosacosB-sinasinB)sina]
=[sinacosacosB+(cosa)^2sinB]/[sinacosacosB-(sina)^2sinB]
=[2sinacosacosB+2(cosa)^2sinB]/[2sinacosacosB-2(sina)^2sinB]
=[sin2acosB+(1+cos2a)sinB]/[sin2acosB+(-1+cos2a)sinB]
=(sin2acosB+cos2asinB+sinB)/(sin2acosB+cos2asinB-sinB)
=[sin(2a+B)+sinB]/[sin(2a+B)-sinB].······④
比较③、④,得:(1+m)/(1-m)=tan(a+B)/tana
∴tan(a+B)=[(1+m)/(1-m)]tana.
注:若需要证明的结论不是我所猜测的那样,则请你补充说明.
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