已知α是第二象限角,且sin(π+α)=k-1/k+1,sin(5π/2+α)=3k-1/k+1,求α得正切值

题目

已知α是第二象限角,且sin(π+α)=k-1/k+1,sin(5π/2+α)=3k-1/k+1,求α得正切值

答案

由诱导公式,sin(π+α)=-sinα=(k-1)/(k+1),即sinα=(1-k)/(k+1)；
sin(5π/2+α)=cosα=(3k-1)/(k+1)
sin²α+cos²α=1；
即：(1-k)²/(k+1)²+(3k-1)²/(k+1)²=1
(1-k)²+(3k-1)²=(k+1)²
10k²-8k+2=k²+2k+1
9k²-10k+1=0
(9k-1)(k-1)
k=1/9 或 k=1
k=1/9时,sinα=(1-k)/(k+1)=4/5,cosα=(3k-1)/(k+1)=-3/5；则tanα=-4/3；且满足是第二象限角.
k=1时,sinα=(1-k)/(k+1)=0,不满足是第二象限角,舍去；
综上,α的正切值是-4/3；
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