How would this equation be simplified?

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AI Thread Summary
The discussion revolves around simplifying the equation that describes how a line reflects off a parabola, initially presented as [(tan[2*arctan(-y/50)]*[1000-(y^2)])/100]+y. The user notes that while the equation works well for visualization, it is cumbersome. They reference Wolfram Alpha, which simplifies the equation to (-1500*y)/([y^2]-2500). Attempts to simplify the original equation using trigonometric identities were initially unsuccessful, but the user eventually realizes that applying the identity for tan(2x) and recognizing the relationship with arctan leads to a clearer understanding. The discussion concludes with a sense of accomplishment in achieving the desired simplification.
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Homework Statement


To satisfy my curiosity, I tried to come up with an equation that describes how a line reflects off a parabola. The equation I came up with is [(tan[2*arctan(-y/50)]*[1000-(y^2)])/100]+y
This equation works wonderfully but its just large and ugly.
The wolfram alpha site for this equation is:
http://www.wolframalpha.com/input/?i=[(tan[2*arctan(-y/50)]*[1000-(y^2)])/100]+y
It just helps visualize the equation. From there, I noticed that wolfram simplifies this down further to (-1500*y)/([y^2]-2500)
How would I go about simplifying the original equation to the nice one wolfram gives?


Homework Equations


[(tan[2*arctan(-y/50)]*[1000-(y^2)])/100]+y


The Attempt at a Solution


I have tried to look at many trig identities involving the tangent functions but none of them seem to help in this case. I have tried moving things around but everything I try just seems to stop with those ugly tangents still left.
 
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You could use the identity:

tan(2x) = 2tanx/ (1- (tanx)^2 )

and tan(arctan x) = x
 
I tried that but on the bottom, you would still be left with 1-(tanx)^2
 
but remember x = arctan(-y/50).
So, 1-(tanx)^2 = 1 - (tan[arctan(-y/50)]^2 = 1 - y^2/2500
 
Yes! I see it now, thanks for your help :)
 
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