已知α是第二象限角,且sin(π+α)=(k-1)/(k+1),sin(5π/2+α)=3k-1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,(1)求α得正切值、余弦值和正切值速求

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

已知α是第二象限角,且sin(π+α)=(k-1)/(k+1),sin(5π/2+α)=3k-1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,(1)求α得正切值、余弦值和正切值速求
已知α是第二象限角,且sin(π+α)=(k-1)/(k+1),sin(5π/2+α)=3k-1
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,(1)求α得正切值、余弦值和正切值
速求

已知α是第二象限角,且sin(π+α)=(k-1)/(k+1),sin(5π/2+α)=3k-1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,(1)求α得正切值、余弦值和正切值速求
sin(π+a)=-sina=(k-1)/(k+1)
sin(5π/2+a)=-cosa=(3k-1)/(k+1)
得:
sina=(k-1)/(k+1)、cosa=(3k-1)/(k+1)
因为:sin²a+cos²a=1
代入,解得:k=1【不符合,舍去】或k=1/9
则:
sina=-4/5、cosa=-3/5
得:tana=sina/cosa=4/3