Question

第19题怎么做 能做多少做多少 急 谢谢

Answer 1

现在？

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嗯

追答

图我就不画了

QWQ

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谢谢

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不客气

Answer 2

用sin^2x+cos^2x=1,加2π和-4
π都可以省略,就是((1-k）/(1+k))^2+((3k-1)/(1+k))^2=1;

Answer 3

Answer 4

sin(2π+α)=sinα=(1-k)/(1+k)
cos(-4π+α)=cos[-(4π-α)]=cos(4π-α)
=cos(2π+2π-α)=cos(2π-α)
=cosα=(3k-1)/(k+1)
∵α是第二象限角
∴sinα>0，cosα<0
∴(1-k)/(k+1)>0，即：(k-1)/(k+1)<0
∴-1<k<1
且(3k-1)/(k+1)<0，则-1<k<1/3
∴-1<k<1/3
∵sin²α + cos²α=1
∴[(1-k)/(k+1)]² + [(3k-1)/(k+1)]²=1
[(1-k)² + (3k-1)]²/(k+1)²=1
(1-2k+k²+9k²-6k+1)/(k+1)²=1
(10k²-8k+2)/(k+1)²=1
∵k+1≠0
∴两边同乘(k+1)²：10k²-8k+2=k²+2k+1
9k²-10k+1=0
(9k-1)(k-1)=0
∴k=1/9或k=1(舍)
∴sinα=(1 - 1/9)/(1 + 1/9)=4/5
cosα=[3•(1/9) - 1]/(1 + 1/9)=-3/5
tanα=sinα/cosα=(4/5)/(-3/5)=-4/3

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