已知函数F(X)=2sin(2x/3+π/6)-1在三角形ABC中,若F(B)=1,且2sin²C-cosC=sin(B-C),求角B与cosC的值

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已知函数F(X)=2sin(2x/3+π/6)-1在三角形ABC中,若F(B)=1,且2sin²C-cosC=sin(B-C),求角B与cosC的值已知函数F(X)=2sin(2x/3+π/

已知函数F(X)=2sin(2x/3+π/6)-1在三角形ABC中,若F(B)=1,且2sin²C-cosC=sin(B-C),求角B与cosC的值
已知函数F(X)=2sin(2x/3+π/6)-1
在三角形ABC中,若F(B)=1,且2sin²C-cosC=sin(B-C),求角B与cosC的值

已知函数F(X)=2sin(2x/3+π/6)-1在三角形ABC中,若F(B)=1,且2sin²C-cosC=sin(B-C),求角B与cosC的值
2sin(2B/3+π/6)-1=1
sin(2B/3+π/6)=1
2B/3+π/6=π/2
B=π/2
2sin²C-cosC=sin(B-C)=cosC
sin²C=cosC=1-cos²C
cos²C+cosC-1=0
cosC=(根号5-1)/2

由F(B)=1得 2sin(2B/3 + π/6) - 1 = 1
推出 sin(2B/3 + π/6) = 1
推出 2B/3 + π/6 = π/2
推出 B = π/2
又由 2(sinC)^2 - cosC = sin(B-C) = sin(π/2 - C) = cosC
推出 1 - (cosC)^2 = (sinC)^2 = cosC
推...

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由F(B)=1得 2sin(2B/3 + π/6) - 1 = 1
推出 sin(2B/3 + π/6) = 1
推出 2B/3 + π/6 = π/2
推出 B = π/2
又由 2(sinC)^2 - cosC = sin(B-C) = sin(π/2 - C) = cosC
推出 1 - (cosC)^2 = (sinC)^2 = cosC
推出 cosC = (根号5 - 1) / 2 或 cosC = (-根号5 - 1) / 2 （因为B是直角所以C一定是锐角，所以负根舍去）

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