已知tanα=2,求1/(sin^2α-sinαcosα-cos^α).由sin^2α=4/5,cos^2α=1/5 且tanα=sinα/cosα=2 可知1/(sin^2α-sinαcosα-cos^α)=1/(sin^2α-3cos^α)=1/(4/5-3/5)=5 想问的是:①sin^2α=4/5,cos^2α=1/5是怎么得出来
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![已知tanα=2,求1/(sin^2α-sinαcosα-cos^α).由sin^2α=4/5,cos^2α=1/5 且tanα=sinα/cosα=2 可知1/(sin^2α-sinαcosα-cos^α)=1/(sin^2α-3cos^α)=1/(4/5-3/5)=5 想问的是:①sin^2α=4/5,cos^2α=1/5是怎么得出来](/uploads/image/z/2496947-59-7.jpg?t=%E5%B7%B2%E7%9F%A5tan%CE%B1%3D2%2C%E6%B1%821%2F%EF%BC%88sin%5E2%CE%B1%EF%BC%8Dsin%CE%B1cos%CE%B1%EF%BC%8Dcos%5E%CE%B1%EF%BC%89.%E7%94%B1sin%5E2%CE%B1%3D4%2F5%2Ccos%5E2%CE%B1%3D1%2F5+%E4%B8%94tan%CE%B1%3Dsin%CE%B1%2Fcos%CE%B1%3D2+%E5%8F%AF%E7%9F%A51%2F%EF%BC%88sin%5E2%CE%B1%EF%BC%8Dsin%CE%B1cos%CE%B1%EF%BC%8Dcos%5E%CE%B1%EF%BC%89%3D1%2F%EF%BC%88sin%5E2%CE%B1%EF%BC%8D3cos%5E%CE%B1%EF%BC%89%3D1%2F%284%2F5%EF%BC%8D3%2F5%29%3D5+%E6%83%B3%E9%97%AE%E7%9A%84%E6%98%AF%EF%BC%9A%E2%91%A0sin%5E2%CE%B1%3D4%2F5%2Ccos%5E2%CE%B1%3D1%2F5%E6%98%AF%E6%80%8E%E4%B9%88%E5%BE%97%E5%87%BA%E6%9D%A5)
已知tanα=2,求1/(sin^2α-sinαcosα-cos^α).由sin^2α=4/5,cos^2α=1/5 且tanα=sinα/cosα=2 可知1/(sin^2α-sinαcosα-cos^α)=1/(sin^2α-3cos^α)=1/(4/5-3/5)=5 想问的是:①sin^2α=4/5,cos^2α=1/5是怎么得出来
已知tanα=2,求1/(sin^2α-sinαcosα-cos^α).
由sin^2α=4/5,cos^2α=1/5 且tanα=sinα/cosα=2 可知1/(sin^2α-sinαcosα-cos^α)=1/(sin^2α-3cos^α)=1/(4/5-3/5)=5 想问的是:①sin^2α=4/5,cos^2α=1/5是怎么得出来的?②1/(sin^2α-sinαcosα-cos^α)是怎么得到1/(sin^2α-3cos^α)的?
已知tanα=2,求1/(sin^2α-sinαcosα-cos^α).由sin^2α=4/5,cos^2α=1/5 且tanα=sinα/cosα=2 可知1/(sin^2α-sinαcosα-cos^α)=1/(sin^2α-3cos^α)=1/(4/5-3/5)=5 想问的是:①sin^2α=4/5,cos^2α=1/5是怎么得出来
①假设α是直角三角形的一个角,因为tanα=2,所以可设一直角边为2,另一直角边为1,所以斜边为根号5,所以sinα=2除以根号5,所以sin^2α=4/5,同理得cos^2α=1/5 ②因为tanα=sinα/cosα=2,所以sinα=2cosα代入即可得
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