Home » Uncategorized » Trigonometric equality

Trigonometric equality

By John Khan in Uncategorized on June 15, 2018.

Prove that in any triangle $ABC$ it holds that

$\sum \sqrt{\frac{\sin A}{\sin B \sin C}} = \sqrt{\frac{2R}{r} \sum \sin A}$

where $R$ denotes the circumradius and $r$ the inradius.

Solution

Using the law of sines we have that

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2 R$

and if we denote $\mathcal{A}$ the area of the triangle then

$\frac{r \left ( a + b + c \right )}{2} = \mathcal{A} = \frac{abc}{4R}$

Thus,

$\begin{align*} \sum \sqrt{\frac{\sin A}{\sin B \sin C}} &= \sum \sqrt{\frac{a}{2R} \cdot \frac{2R}{b} \cdot \frac{2R}{c}} \\ &=\sum \sqrt{\frac{ a}{bc} \cdot 2 R} \\ &=\sum \sqrt{\frac{a}{bc} \cdot \frac{abc}{2\mathcal{A}}} \\ &=\frac{a+b+c}{\sqrt{2 \mathcal{A}}} \\ &= \sqrt{\frac{a+b+c}{\frac{2\mathcal{A}}{a+b+c}}} \\ &= \sqrt{\frac{a+b+c}{r}} \\ &= \sqrt{\frac{1}{r} \sum a} \\ &= \sqrt{\frac{2R}{r} \sum \frac{a}{2R}} \\ &= \sqrt{\frac{2R}{r} \sum \sin A} \end{align*}$