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Apr 13, 2017 at 12:55 history edited CommunityBot
replaced http://mathematica.stackexchange.com/ with https://mathematica.stackexchange.com/
Jun 15, 2012 at 14:14 comment added Ben Allgeier Oh yeah, I meant to take the square root above. Thanks.
Jun 15, 2012 at 13:06 comment added J. M.'s missing motivation @Ben, no; $\mathrm{hav}(x)=\sin^2\dfrac{x}{2}$. Otherwise, yes; I'm exploiting the fact that $\mathrm{hav}(x)$ is not immediately simplified to more basic trigonometric functions.
Jun 15, 2012 at 12:55 comment added Ben Allgeier So what really makes this work? Mathematically, Haversine[2x] is equal to Sin[x] on the interval [0, Pi/2]. It looks like plt has the form {Line[], Line[] }. So the plot thinks of it as two different functions still. Is this right? Interesting solution.
Jun 14, 2012 at 22:50 history answered J. M.'s missing motivation CC BY-SA 3.0