已知asin(O+a)=bsin(O+B),求证:tanO=bsinB-asina/acosa-bcosB
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已知asin(O+a)=bsin(O+B),求证:tanO=bsinB-asina/acosa-bcosB
已知asin(O+a)=bsin(O+B),求证:tanO=bsinB-asina/acosa-bcosB
已知asin(O+a)=bsin(O+B),求证:tanO=bsinB-asina/acosa-bcosB
asin(O+a)=bsin(O+B)
a(sinOcosa+cosOsina)=b(sinOcosB+cosOsinB)
a(tanOcosa+sina)=b(tanOcosB+sinB)
atanOcosa+asina=btanOcosB+bsinB
(acosa-bcosB)tanO=bsinB-asina
tanO=bsinB-asina/acosa-bcosB
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