sin(5π/12-α)=1/3化简
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COS(π/12-α) - √3cos(5π/12+α)
= 2 [(1/2)COS(π/12-α) - (√3/2)罪(π/12 -α)](∵COS(5π/12+α)= SIN [π/2-(5π/12+α)] = SIN(π/12-α))
= 2sin [π/6- (π/12-α)]
= 2sin(π/12+α)
= 2 [(1/2)COS(π/12-α) - (√3/2)罪(π/12 -α)](∵COS(5π/12+α)= SIN [π/2-(5π/12+α)] = SIN(π/12-α))
= 2sin [π/6- (π/12-α)]
= 2sin(π/12+α)
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