求证:sin(kπ-α)cos(kπ+α)/sin[(k+1)π+α]cos[(k+1)π+α]=-1，k∈z
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解决时间 2021-03-17 12:24
- 提问者网友：几叶到寒
- 2021-03-16 16:57
求证:sin(kπ-α)cos(kπ+α)/sin[(k+1)π+α]cos[(k+1)π+α]=-1，k∈z
最佳答案
- 五星知识达人网友：几近狂妄
- 2021-03-16 17:30
原题是:求证(sin(kπ-α)cos(kπ+α))/(sin((k+1)π+α)cos((k+1)π+α))=-1，k∈Z.
证明:设A=sin(kπ-α)cos(kπ+α)，B=sin((k+1)π+α)cos((k+1)π+α)
2A=sin((kπ-α)+(kπ+α))+sin((kπ-α)-(kπ+α))
=sin(2kπ)+sin(-2α)
=-sin2α
2B=sin(2((k+1)π+α))
=sin(2(k+1)π+2α)
=sin2α
左边=A/B=(2A)/(2B)=(-sin2α)/sin2α=-1=右边
所以 (sin(kπ-α)cos(kπ+α))/(sin((k+1)π+α)cos((k+1)π+α))=-1，k∈Z.
希望能帮到你！
证明:设A=sin(kπ-α)cos(kπ+α)，B=sin((k+1)π+α)cos((k+1)π+α)
2A=sin((kπ-α)+(kπ+α))+sin((kπ-α)-(kπ+α))
=sin(2kπ)+sin(-2α)
=-sin2α
2B=sin(2((k+1)π+α))
=sin(2(k+1)π+2α)
=sin2α
左边=A/B=(2A)/(2B)=(-sin2α)/sin2α=-1=右边
所以 (sin(kπ-α)cos(kπ+α))/(sin((k+1)π+α)cos((k+1)π+α))=-1，k∈Z.
希望能帮到你！
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