maggie1120
幼苗
共回答了20个问题采纳率:85% 举报
方法一:
1、当k为偶数时,设k=2n,其中n为整数.此时,
原式=cos[(6n+1)π/3+α]+cos[(6π-1)π/3-α]
=cos(π/3+α)+cos(-π/3-α)
=cos(π/3+α)+cos(π/3+α)
=2cos(π/3+α)
=2[cos(π/3)cosα-sin(π/3)sinα]
=2cos(π/3)cosα-2sin(π/3)sinα
=2(1/2)cosα-2(√3/2)sinα
=cosα-√3sinα.
2、当k为奇数时,设k=2n+1,其中n为整数.此时,
原式=cos[(6n+4)π/3+α]+cos[(6π-4)π/3-α]
=cos(4π/3+α)+cos(-4π/3-α)
=cos(4π/3+α)+cos(4π/3+α)
=2cos(4π/3+α)
=-2cos(π/3+α)
=√3sinα-cosα.
综上1、2所述,当k为偶数时,原式=cosα-√3sinα,当k为奇数时,原式=√3sinα-cosα.
方法二:
原式=cos[kπ+(π/3+α)]+cos[kπ-(π/3+α)]
=coskπcos(π/3+α)-sinkπsin(π/3+α)+coskπcos(π/3+α)+sinkπsin(π/3+α)
=2coskπcos(π/3+α)
=coskπ(cosα-√3sinα)
显然,当k为偶数时,coskπ=1,此时原式=cosα-√3sinα,
当k为奇数时,coskπ=-1,此时原式=-(cosα-√3sinα)=√3sinα-cosα.
1年前
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