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cos(π/12-α)-√3cos(5π/12+α)
=2[(1/2)cos(π/12-α)-(√3/2)sin(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)sin(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)
1年前
6
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你能帮帮他们吗