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)](/uploads/image/z/1669801-49-1.jpg?t=%E5%8C%96%E7%AE%80+sin%7B%5B%284n%2B1%29%CF%80%2F4%5D%2B%CE%B1%7D%2Bcos%7B%5B%284n-1%29%CF%80%2F4%5D-%CE%B1%7D%2C%EF%BC%88n%E2%88%88Z%EF%BC%89)
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
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)]
...
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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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