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Interesting Identity 
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#1
Mar912, 01:44 PM

P: 1

Hey guys,
I stumbled on an interesting and unexpected identity when looking for a simpler summation technique for inverse kinementics. Basically, I was trying to find a simpler way of summing IK vectors for some particular armature (like a tentacle or multibranch armature on a robot). This question doesn't have to do with IK, but I accidentally calculated an identity rule I've not been able to attach to any other identity rule in trigonometry. This is the XPlore code that graphs the identity: [The trig set is in degrees, not radians] My question is: Is there an existing trig identity or rule (or set of iedntities or rules) that explains why the above functional set works? Basically the "law" that it asserts is: It's not an important question, just something that's been bugging me for a while. Thanks for looking, CM 


#2
Mar912, 05:14 PM

P: 688

Also, to avoid confusing the two meanings of 'x', you may want to express the lefthand side as sum[t=0 to x](sin(t)) ; 'x' is the limit of the sum, not really the argument to sin(). The x's in the righthand side are correct, as they mean the same value of 'x' as the limit of the sum in the lefthand side. So, if I get you right, the conjecture is[tex]\sum_{t=0}^x \sin t = \left( \sum_{t=0}^{180} \sin t \right) \sin^2(x/2) + (\sin x)/2[/tex]with the arguments in degrees, not radians. The two overlapping curves (lefthand side, righthand side) looking like this, for x=0 to 180: 


#3
Mar912, 05:21 PM

P: 14

Hi,
Let [itex]x \in \mathbb R, n \in \mathbb N[/itex]. I'll be using radians instead of degrees, so I'll show that we have : [tex]\sum_{k=0}^n \sin k = f(\pi) \sin^2\left(\dfrac{x}{2}\right)+\dfrac{\sin x}{2}[/tex] Where : [itex]f(n) = \displaystyle\sum_{k=0}^n \sin k[/itex] (NB: that expression isn't yet defined on all real numbers and more particularly on [itex]n=\pi[/itex], but it will be as soon as we get another expression of it). We have : [tex]f(n) = \sum_{k=0}^n \text{Im}(e^{ik}) = \text{Im}\left(\sum_{k=0}^n e^{ik}\right) = \text{Im}\left(\sum_{k=0}^n (e^{i})^k\right)[/tex] And since [itex]e^i \neq 1[/itex], applying the formula for the sum of a geometric sequence gives : [tex]f(n) = \text{Im}\left(\dfrac{(e^i)^{n+1}1}{e^i1}\right) = \text{Im}\left(\dfrac{e^{i(n+1)}1}{e^i1}\right)[/tex] By applying half angle formulas : [tex]f(n) = \text{Im}\left(\dfrac{e^{i(n+1)/2}\cdot 2i\sin\left(\dfrac{n+1}{2}\right)}{e^{i/2}\cdot 2i\sin(1/2)}\right)[/tex] After simplifying we finally get : [tex]f(n) = \dfrac{\sin\left(\dfrac{n+1}{2}\right) \sin\left(\dfrac{n}{2}\right)}{\sin\left(\dfrac{1}{2}\right)}[/tex] That's a quite "simple" expression that only uses products, that naturally extends [itex]f(n)[/itex] to real numbers. I'll now show that the quantity [itex]f(x)[/itex] is exactly equal to [itex]f(\pi) \sin^2\left(\dfrac{x}{2}\right) + \dfrac{\sin x}{2}[/itex] for all [itex]x \in \mathbb R[/itex]. According to the expression of [itex]f(x)[/itex] obtained above, we have : [tex]f(\pi) = \dfrac{\sin\left(\dfrac{\pi+1}{2}\right) }{\sin\left(\dfrac{1}{2}\right)} = \dfrac{\cos\left(\dfrac{1}{2}\right)}{\sin\left( \dfrac{1}{2}\right)}[/tex] Thus : [tex]f(\pi) \sin^2\left(\dfrac{x}{2}\right) + \dfrac{\sin x}{2} = \dfrac{\cos\left(\dfrac{1}{2}\right)\sin^2\left( \dfrac{x}{2}\right) }{\sin\left( \dfrac{1}{2}\right)}+\dfrac{\sin x}{2} = \\ \dfrac{2\cos\left( \dfrac{1}{2}\right)\sin^2\left( \dfrac{x}{2}\right) + \sin\left( \dfrac{1}{2}\right)\left[2\sin\left(\dfrac{x}{2}\right)\cos\left( \dfrac{x}{2}\right)\right]}{2\sin\left( \dfrac{1}{2}\right)} = \\\dfrac{\sin\left(\dfrac{x}{2}\right)\left[\cos\left(\dfrac{1}{2}\right) \sin\left(\dfrac{x}{2}\right) + \sin\left(\dfrac{1}{2}\right)\cos\left(\dfrac{x}{2}\right)\right]}{\sin\left(\dfrac{1}{2}\right)} = \\\dfrac{\sin\left(\dfrac{x}{2}\right) \sin\left(\dfrac{x+1}{2}\right)}{\sin\left(\dfrac{1}{2}\right)} = f(x)[/tex] You just need to change from radians to degrees and it's done. 


#4
Mar912, 05:40 PM

P: 688

Interesting Identity
Edit: On second thought, the technique is beautiful but there has to be an error here. [itex]\sum_{k=0}^n \text{Im}(e^{ik})[/itex] is not the same as [itex]\sum_{k=0}^n \text{Im}(e^{ik \pi/180})[/itex], which was what the OP had in mind, regardless of whether you later extend the definition for a real n: "mind the step" ! Edit2: Ah, you already said that. "You just need to change from radians to degrees and it's done."  i.e. Repeat the proof, this time with a pi/180 multiplier. When will I learn to read posts to the end. 


#5
Mar1012, 03:11 AM

P: 688

So, if I followed right, you end up with[tex]
f(n) = \frac{\sin\left(\frac{\pi n}{360}\right) \sin\left(\frac{\pi(n+1)}{360}\right)} {\sin\left(\frac{\pi}{360}\right)} [/tex]which, after expanding the sum of sines, and using a halfangle formula again, ends up in[tex] f(n) = \frac{\sin^2\left(\frac{\pi n}{360}\right)}{\tan\left(\frac{\pi}{360}\right)} + \frac{\sin\left(\frac{\pi n}{180}\right)}2 [/tex]so that[tex] f(180) = \frac 1 {\tan\left(\frac{\pi}{360}\right)} [/tex]from which the conjecture[tex] f(n) = f(180) \sin^2(\pi n/360) + (\sin \pi n/180)/2 [/tex]follows. 


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