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This diagram:

enter image description here

indicates how the expression $2r\sin((\Theta_2 - \Theta_1)/2)$ for chord of circle of radius r subtending angle $\Theta_2 - \Theta_1$ can be validated geometrically (see that the segment n is half of $\delta$, where $\delta$ is $d(P1, P2))$, but I am having trouble coming up with the algebraic statement using the distance formula as a starting point and the coordinates $P2(r \cos\Theta_2, r\sin \Theta_2)$ and P1 similarly defined.

Using the distance formula I get $\delta = r \sqrt{2 - 2\cos(\Theta_2 - \Theta_1)}$ straightforwardly (using pythagorean theorem twice to get the leading '2' within the radical, and the cosine difference formula for the cosine expression also within the radical).

Bernard
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1 Answers1

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I don't know if this answers your question.

For a circle with center at the origin, radius vector in polar coordinates but cartesian basis is $$ \mathbf R = R(\cos\theta, \sin\theta)^T $$ For the chord, $\mathbf L$ we have $$ \mathbf L = \mathbf R_1 - \mathbf R_2 \\ = R(\cos\theta_1 -\cos\theta_2, \sin\theta_1 -\sin\theta_2)^T $$ The square of its length is $$ L^2 = \mathbf L\cdot \mathbf L = R^2 ((\cos\theta_1 -\cos\theta_2)^2 + (\sin\theta_1 -\sin\theta_2)^2) \\ = R^2(2 - 2(\cos\theta_1\cos\theta_1+\sin\theta-1\sin\theta_2)) \\ = 2R^2(1 - (\cos(\theta_1-\theta_2))) \\ $$

Physor
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  • This looks like another way to derive the expression in my last paragraph (which I did from distance formula in cartesian coords - noting I had a sign difference from yours). But I don't see how you've shown the algebraic connection with $2r sin ((\Theta_2 - \Theta_1)/2)$ - that geometrically is seen the same from diagram (n is $sin ((\Theta_2 - \Theta_1)/2$, Euclidean geom shows n equal to half the distance between P1, P2)? –  Oct 05 '20 at 23:17
  • It is now corrected, thanks for the comment! – Physor Oct 06 '20 at 08:23