求证:sin(kπ-α)cos(kπ+α)⼀sin[(k+1)π+α]cos[(k+1)π+α]=-1,k∈z

2026年09月22日 15:30
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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.

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