已知sin(π/3-X),则cos(5π/6-X)=?

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已知sin(π/3-X),则cos(5π/6-X)=?已知sin(π/3-X),则cos(5π/6-X)=?已知sin(π/3-X),则cos(5π/6-X)=?cos(5π/6-X)=-cos[π-

已知sin(π/3-X),则cos(5π/6-X)=?
已知sin(π/3-X),则cos(5π/6-X)=?

已知sin(π/3-X),则cos(5π/6-X)=?
cos(5π/6-X)
=-cos[π-(5π/6-X)]
=-cos(x+π/6)
=-sin[π/2-(x+π/6)]
=-sin(π/3-X)

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