化简 sin{[(4n+1)π/4]+α}+cos{[(4n-1)π/4]-α},（n∈Z）

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化简 sin{[(4n+1)π/4]+α}+cos{[(4n-1)π/4]-α},（n∈Z）
化简 sin{[(4n+1)π/4]+α}+cos{[(4n-1)π/4]-α},（n∈Z）

化简 sin{[(4n+1)π/4]+α}+cos{[(4n-1)π/4]-α},（n∈Z）
sin{[(4n+1)π/4]+α}+cos{[(4n-1)π/4]-α}
=sin{nπ+[(π/4)+α]}+cos{nπ-[(π/4)+α]}
=sin(nπ)cos(π/4+α)+cos(nπ)sin(π/4+α)+cos(nπ)cos(π/4+α)+sin(nπ)sin(π/4+α)]
=cos(nπ)sin(π/4+α)+cos(nπ)cos(π/4+α) //sin(nπ)=0
=cos(nπ)[sin(π/4+α)+cos(π/4+α)]
=±[sin(π/4)cosα+cos(π/4)sinα+cos(π/4)cosα-sin(π/4)sinα] //cos(nπ)=±1
=±(√2)cosα //cos(π/4)=sin(π/4)=(√2)/2

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sin{[(4n+1)π/4]+α}+cos{[(4n-1)π/4]-α}
=sin[(nπ+π/4)+α]+cos[(nπ-π/4)-α]
=sin[nπ+(π/4+α)]+cos[nπ+π/2-π/4-α]
=sin[nπ+(π/4+α)]+cos[nπ+π/2-(π/4+α)]
=sin[nπ+(π/4+α)]+cos[π/2-(π/4+α-nπ)]
=sin(nπ+π/4+α)+sin(π/4+α-nπ)
=sin(nπ+π/4+α)-sin(nπ-π/4-α)
=2cos[(nπ+π/4+α+nπ-π/4-α)/2] sin[(nπ+π/4+α-nπ+π/4+α)/2]
=2cos(nπ) sin(π/4+α)
=±2sin(π/4+α)

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