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The other day I came across an identity in the book "Problems from the Book" and it was presented as well known:

$$\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} = \frac{r}{4R}$$

However I wasn't familiar with thee identity so I tried proving it.

Here is part of my attempt in solving this problem:

I first rewrote $\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}$ as $\frac{r^3}{AI\cdot BI \cdot CI}$ (where I is the incenter of $\bigtriangleup ABC$). Then I tried using the fact that [ABC] =abc/4R (where [ABC] represents the area of $\bigtriangleup ABC$)which allowed me to rewrite the equation as $\frac{r^2}{AI\cdot BI \cdot CI} = \frac{[ABC]}{abc}$. I tried various things at this point but none of my attempts were successful...

Does anyone have a proof of this identity?

Rio
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  • Are you familiar with the half-angle formulae in terms of $a,b,c,s$? – Fawkes4494d3 Feb 09 '21 at 03:06
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    You need to add a definition for what $A, B, C, r$ and $R$ are. Otherwise this question is impossible to answer. – jMdA Feb 09 '21 at 03:06
  • It is interesting to note that each of the inradius, and the exradii, $r,r_a,r_b,r_c$ all have similar formulae, with $r_a$ having two external angles bisected in $B,C$, so that we will have $$r_a = 4R\sin\left( \dfrac{A}2 \right)\sin\left( \dfrac{\pi - B}2 \right)\sin\left( \dfrac{\pi - C}2 \right) \ = 4R\sin\left( \dfrac{A}2 \right)\cos\left( \dfrac{B}2 \right)\cos\left( \dfrac{C}2 \right)$$ – Fawkes4494d3 Feb 09 '21 at 03:09
  • https://math.stackexchange.com/questions/734395/how-to-prove-that-fracrr1-cos-a-cos-b-cos-c and https://math.stackexchange.com/questions/176892/prove-trigonometry-identity-for-cos-a-cos-b-cos-c – lab bhattacharjee Feb 09 '21 at 05:06

1 Answers1

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$$LHS = \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} $$

$$= \sqrt{\frac{(s-b)(s-c)}{bc}} \cdot \sqrt{\frac{(s-c)(s-a)}{ca}} \cdot \sqrt{\frac{(s-a)(s-b)}{ab}}$$

$$= \frac{(s-a)(s-b)(s-c)}{abc}$$

Note that $$r = \frac{\Delta}{s}$$ $$R = \frac{abc}{4 \Delta}$$ $$\Delta^2 = s(s-a)(s-b)(s-c)$$ Now $$ RHS = \frac{r}{4R} = \frac{\Delta}{s} \cdot \frac{\Delta}{abc} = \frac{(s-a)(s-b)(s-c)}{abc}$$

Hence LHS = RHS

PTDS
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