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I'm trying to prove the following formula.

Given a spherical triangle with side lengths $a,b,c$ and interior angles $ \alpha,\beta,\gamma $ prove the following formula $$2\sin \frac{A}{2} =\frac{\sqrt{\sin(s) \sin(s-a) \sin(s-b) \sin(s-c)}}{\cos \frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2}}$$ where $A= \alpha+\beta+\gamma -\pi$ and $s=\frac{1}{2}(a+b+c)$.

I wanted to show that $$2 \sin \frac{A}{2}\cos \frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2} = \sqrt {\sin(s) \sin(s-a)\sin(s-b)\sin(s-c)} .$$

So far I was able to prove that

$$\sqrt {\sin(s) \sin(s-a)\sin(s-b)\sin(s-c)} = \sin(a)\sin(b)\sin(\gamma)$$

$$\sin(a)=2\sin\frac{a}{2} \cos\frac{a}{2}$$

$$\sin(b)=2\sin\frac{b}{2} \cos\frac{b}{2}$$

so I get

$$\sqrt {\sin(s) \sin(s-a)\sin(s-b)\sin(s-c)} = 2\sin\frac{a}{2} \cos\frac{a}{2} \cdot 2\sin\frac{b}{2} \cos\frac{b}{2} \cdot \sin(\gamma)$$ $$ = 4\sin\frac{a}{2}\sin\frac{b}{2}\sin(\gamma)\cos\frac{a}{2}\cos\frac{b}{2}.$$

Now I only have to show that $2\sin \frac{A}{2}\cos\frac{c}{2} =4\sin\frac{a}{2}\sin\frac{b}{2}\sin(\gamma)$, or equivalently $\sin \frac{A}{2}\cos\frac{c}{2} =2\sin\frac{a}{2}\sin\frac{b}{2}\sin(\gamma)$.

Since I know that $A-(\alpha+\beta)+\pi= \gamma $, it follows $$2\sin\frac{a}{2}\sin\frac{b}{2}\sin(\gamma)=2\sin\frac{a}{2}\sin\frac{b}{2}\sin(A-(\alpha+\beta)+\pi).$$ But I have no idea what to do now.

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Polymorph
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    That looks like a spherical version of Heron's formula. FWIW, the usual version, aka L'Huilier's formula uses tangents, as shown in https://math.stackexchange.com/a/66731/207316 – PM 2Ring Jan 06 '18 at 15:25
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    Your $A$ is usually denoted by $E$ for spherical excess. – Somos Jan 06 '18 at 15:27

1 Answers1

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According to Wiki Half angle formulae , we have $$ \begin{align*} \sin{\frac{\alpha}{2}}&= \sqrt{\frac{\sin(s-b)\sin(s-c)}{\sin b\sin c}},\\ \cos{\frac{\alpha}{2}} &= \sqrt{\frac{\sin s\sin(s-a)}{\sin b\sin c}}. \end{align*} $$ Then we can use the angle sum identities to expand $\sin \frac{A}{2}=-\cos\left(\frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}\right)$ as follows $$ \begin{align*} \sin \frac{A}{2} &=\sin \frac{\alpha+\beta+\gamma -\pi}{2}\\ &=-\cos\left(\frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}\right)\\ &= \sin\frac{\alpha}{2}\sin\frac{\beta}{2}\cos\frac{\gamma}{2} + \sin\frac{\alpha}{2}\cos\frac{\beta}{2}\sin\frac{\gamma}{2} + \cos\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}-\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2} \\ &=\frac{\sin(s-a)+\sin(s-b)+\sin(s-c)-\sin(s)}{\sin a\sin b\sin c}\sqrt{\sin(s) \sin(s-a) \sin(s-b) \sin(s-c)}\\ &=\frac{4\sin\frac{a}{2}\sin\frac{b}{2}\sin\frac{c}{2}}{\sin a\sin b\sin c}\sqrt{\sin(s) \sin(s-a) \sin(s-b) \sin(s-c)}\\ &=\frac{4\sin\frac{a}{2}\sin\frac{b}{2}\sin\frac{c}{2}}{8\sin \frac{a}{2}\cos \frac{a}{2}\sin \frac{b}{2}\cos\frac{b}{2}\sin \frac{c}{2}\cos\frac{c}{2}}\sqrt{\sin(s) \sin(s-a) \sin(s-b) \sin(s-c)}\\ &=\frac{\sqrt{\sin(s) \sin(s-a) \sin(s-b) \sin(s-c)}}{2\cos \frac{a}{2}\cos\frac{b}{2}\cos\frac{c}{2}}, \end{align*} $$ where the trigonometric identity $$ \begin{align*} \sin(s-a)+\sin(s-b)+\sin(s-c)-\sin(s)&=\sin\frac{b+c-a}{2}+\sin\frac{a+c-b}{2}+\sin\frac{a+b-c}{2}-\sin\frac{a+b+c}{2}\\ &=4\sin\frac{a}{2}\sin\frac{b}{2}\sin\frac{c}{2} \end{align*} $$ can be proved by rewriting sine in terms of exponentials $$ \sin (x)=\frac{e^{i x}-e^{-i x}}{2 i}. $$

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