I'm trying to prove the following formula.

Given a spherical triangle with side lengths 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

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

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.

  • $\begingroup$ That looks like a spherical version of Heron's formula. FWIW, the usual version, aka L'Huilier's formula uses tangents, as shown in math.stackexchange.com/a/66731/207316 $\endgroup$ – PM 2Ring Jan 6 '18 at 15:25
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    $\begingroup$ Your $A$ is usually denoted by $E$ for spherical excess. $\endgroup$ – Somos Jan 6 '18 at 15:27

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