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The equation sin^2theta=(x^2+y^2)/(2x y)...

The equation `sin^2theta=(x^2+y^2)/(2x y),x , y!=0` is possible if

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`sin^2theta=(x^2+y^2)/(2xy)`
`(x^2+y^2)/(2xy)=1/2(x^2/(xy)+y^2/(xy))`
`=1/2[x/y+y/x]`
`=1/2[sqrt(x/y)-sqrt(y/x))^2+2]`
`sin^2theta<=1`
`(x^2+y^2)/(2xy)=1`
`x^2+y^2=2xy`
`x^2+y^2-2xy=0`
...
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