A Math Problem~FOR 已知A∈(-π/2,π/2),lg(1+sinA)=m,lg(1/(1-sinA))=n求lgcosA
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A Math Problem~FOR 已知A∈(-π/2,π/2),lg(1+sinA)=m,lg(1/(1-sinA))=n求lgcosA
A Math Problem~FOR
已知A∈(-π/2,π/2),lg(1+sinA)=m,lg(1/(1-sinA))=n
求lgcosA
A Math Problem~FOR 已知A∈(-π/2,π/2),lg(1+sinA)=m,lg(1/(1-sinA))=n求lgcosA
lg(1+sinA)=m,lg(1/(1-sinA))=n
lg(1/(1-sinA))=-lg(1-sinA)=n
lg(1-sinA)=n
lg(1+sinA)+lg(1-sinA)
=lg[(1-sinA)(1+sinA)]
=lg[(cosA)^2]
=2lgcosA=m-n
lgcosA=(m-n)/2
lg(1+sinA)-lg(1/(1-sinA))=lg((1+sinA)*(1-sinA))=
lg(1-sinA的平方)=lg(cosA的平方)=2lgcosA=m-n
故lgcosA=m-n/2
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