11_hwc1SolnsODDA

11_hwc1SolnsODDA - ² , which is the form we want. In the...

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Next, since the triangle with vertices O , A and C is a right triangle with unit hypotenuse, C is the point (cos(2 ϕ ) , sin(2 ϕ )). Using the double angle formulas, this gives us C = (cos 2 ( ϕ ) - sin 2 ( ϕ ) , 2 sin( ϕ ) cos( ϕ )) , and hence f ( e 1 ) = ± cos 2 ( ϕ ) - sin 2 ( ϕ ) 2 sin( ϕ ) cos( ϕ ) ² . ( * * * ) All we have to do now is to express this in terms of the slope s instead of the angle ϕ . By the definitions of the slope and the tangent, s and ϕ are related by tan( ϕ ) = s . Therefore, by simple trigonometry, cos( ϕ ) = 1 1 + s 2 and sin( ϕ ) = s 1 + s 2 . Hence we can rewrite ( * * * ) as f ( e 1 ) = 1 1 + s 2 ± 1 - s 2 2 s
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Unformatted text preview: ² , which is the form we want. In the same way one sees, f ( e 2 ) = 1 1 + s 2 ± 2 s s 2-1 ² , and hence, from ( ** ), A f = 1 1 + s 2 ± 1-s 2 2 s 2 s s 2-1 ² . It is much simpler to find A g , since clearly g ( e 1 ) = e 1 and g ( e 2 ) =-e 2 , and so A g = ± 1-1 ² . Hence, from ( * ), A g ◦ f = A g A f = 1 1 + s 2 ± 1-1 ²± 1-s 2 2 s 2 s s 2-1 ² = 1 1 + s 2 ± 1-s 2 2 s-2 s 1-s 2 ² . 21 /september/ 2005; 22:09 12...
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This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.

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