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ABC is an equilateral spherical triangle in which small displacements are made, in the sides and angles, of such a nature that the triangle remains equilateral. Prove that $$ \frac{da}{dA} = \cos\left({\frac{A}{2}}\right) \cot\left({\frac{a}{2}}\right) $$

I would have mentioned my progress. But all I have been able to done is differentiate different trigonomeric formulae to get an expression which is of no use to simplify, since it never matches with answer on a numerical verification.

Please help. A hint on how to start might be enough.

NOTE: This question is Q9 of Execrcise 1 from W.M. Smart's Spherical Astronomy.

amWhy
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Cheeku
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  • what is $a,A$ ? Is there a picture in your book? – sigmatau May 18 '13 at 13:05
  • @AmireBendjeddou a is any of the sides and A is any of the angles;the standard symbols.

    No, no pictures.

     – Cheeku May 18 '13 at 13:06
  • with spherical triangle do you mean a triangle inscribed in a circle with one side beeing an arc of it? – sigmatau May 18 '13 at 13:07
  • @AmireBendjeddou No, with spherical triangle I mean a 'spherical triangle'. Just Google it. – Cheeku May 18 '13 at 13:08
  • Is this the formulation from the book? It’s a very sloppy way of stating the problem, which just involves the relation between the side-length $s$ and the angle $\theta$ of an equiangular sph. triangle. Have you bisected the triangle to get a right triangle with hypotenuse $s$ and one leg $s/2$, and angles $\theta$ and $\theta/2$? – Lubin May 18 '13 at 14:48

2 Answers2

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I don’t believe the formula is correct as quoted.

We’re dealing with an equilateral (and equiangular!) spherical triangle, and to avoid (my own) confusion, I’ll call the side-length $s$ and the angle $\theta$. Both are angles, of course. The most obvious thing is to bisect the triangle into two right triangles, each with hypotenuse $s$ and one side $s/2$, this latter side being opposite the angle $\theta/2$, while the other angle of the triangle is still $\theta$.

Now, one of the standard formulas for right spherical triangles is $$ \cos c=\cot A\cot B\,, $$ and for us, $c=s$, $A=\theta$, $B=\theta/2$. So we differentiate our relation $\cos s=\cot\theta\cot(\theta/2)$, and get \begin{align} -\sin s\frac{ds}{d\theta}&=-\cot\theta\csc^2(\theta/2)\cdot\frac12 -\csc^2\theta\cot(\theta/2)\\ \frac{ds}{d\theta}&=\frac{\frac12\cot\theta\csc^2(\theta/2)+\csc^2\theta\cot(\theta/2)}{\sin s}\,. \end{align} Now let’s test this out on the nicest possible equilateral triangle, the one whose sides and angles are both $90^\circ$. Here, the first term in the numerator drops out, ’cause $\cot90^\circ=0$, but everything else evaluates to $1$, and so even $ds/d\theta=1$ at this test-point. Yet, the formula evaluates to $\sqrt2/2$, unequal to $1$.

Lubin
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  • Actually, even I got it as $\cot{\frac{a}{2}} \cot{\frac{A}{2}}$, which in the example you gave does evaluate to 1.

    Thanks! The probability of the relation being wrongly quoted rises...

     – Cheeku May 19 '13 at 02:54
  • I’d love to see how you derived the correct formula, as I guess you did. I couldn’t do anything with the complicated expression above to simplify it. – Lubin May 19 '13 at 20:02
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On page 7 there is a "fundamental formula" that for an equilateral triangle is

$$ \cos a = \cos^2 a + (\sin^2 a) \cos A$$

The rest is a differential calculus problem of implicit differentiation on the equation to write a relation between $da$ and $dA$ and thence to calculate $\frac{da}{dA}$; followed by a (plane) trigonometry problem of using half- or double-angle formulas to compare the formula for $\frac{da}{dA}$ from differentiation to the formula given in the question.

One reason to use the half-angle expressions is that if the $a$ and $A$ terms are moved to opposite sides of the equation, the $A$ side is well behaved, just a cosine, but on the $a$ side there is a removable singularity at $\sin a = 0$, that can be thought of as coming from factors of $\sin^2 (a/2)$ in the numerator and denominator. It could be easier or more illuminating to first remove those factors and then do the implicit differentiation.

zyx
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  • So one gets $\cos a 2 \sin^2 (a/2) = 4 \sin^2 (a/2) \cos^2(a/2) \cos A$ or $\cos a = 2 \cos^2(a/2) \cos A$ or $\cos a = (1 + \cos a) \cos A$. Implicit differentiation gives $-\sin a (da) = \cos A (-\sin a) (da) + (1 + \cos a) (-\sin A)(dA)$ so that $da (1 - \cos A) (\sin a) = dA (\sin A) (1 + \cos a)$ or $\sin^2 (A/2) \sin a da = \cos^2 (a/2) \sin A dA$ or $ cot(a/2) da = cot(A/2) dA$. This says $\log \sin (a/2)$ and $\log \sin (A/2)$ differ by a constant for equilateral spherical triangles, or $\sin (a/2)/\sin (A/2)$ is constant . That looks like a spherical law of sines. – zyx May 21 '13 at 22:52
  • All of this is fine. I have tried this. But in the end, you still don't get the expression claimed by the book...:) – Cheeku May 23 '13 at 01:34