若sina=1/3,求cos((2k+1/2)π+a)+cos((2k-1/2)π-a)cos[(2k+1)π/2+a]+cos[(2k-1)π/2-a]

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若sina=1/3,求cos((2k+1/2)π+a)+cos((2k-1/2)π-a)cos[(2k+1)π/2+a]+cos[(2k-1)π/2-a]若sina=1/3,求cos((2k+1/2)

若sina=1/3,求cos((2k+1/2)π+a)+cos((2k-1/2)π-a)cos[(2k+1)π/2+a]+cos[(2k-1)π/2-a]
若sina=1/3,求cos((2k+1/2)π+a)+cos((2k-1/2)π-a)
cos[(2k+1)π/2+a]+cos[(2k-1)π/2-a]

若sina=1/3,求cos((2k+1/2)π+a)+cos((2k-1/2)π-a)cos[(2k+1)π/2+a]+cos[(2k-1)π/2-a]
答:
sina=1/3
cos[(2k+1/2)π+a]+cos[(2k-1/2)π-a]
=cos(2kπ+π/2+a)+cos(2kπ-π/2-a)
=cos(π/2+a)+cos(-π/2-a)
=cos[π-(π/2-a)]+cos(π/2+a)
=-2cos(π/2-a)
=-2sina
=-2/3

cos[(2k+1/2)π+a]+cos[(2k-1/2)π-a]
=cos[(2kπ+π/2+a]+cos[(2kπ-π/2-a]
=cos(π/2+a)+cos(π/2+a)
=-2sina=-2/3cos[(2k+1/2)π+a]+cos[(2k-1/2)π-a]
=cos[(2kπ+π/2+a]+cos[(2kπ-π/2-a]
应该是=cos[(kπ...

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cos[(2k+1/2)π+a]+cos[(2k-1/2)π-a]
=cos[(2kπ+π/2+a]+cos[(2kπ-π/2-a]
=cos(π/2+a)+cos(π/2+a)
=-2sina=-2/3

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