Trigonometry-circumcircle and sides of triangle.

Question

How to prove that $$4R\sin A\sin B\sin C=a \cos A+b \cos B+c\cos C$$ where R is the radius of the circumcircle and $a$,$b$ and $c$ the respective sides of the triangle.

I wrote $R=a/2\sin A$ and then got $\sin A$ cancelled then changed $\sin B \sin C$ to the sums of cosine but that didn't work.

It looks like that you ask those questions that you aren't able to do in your first try. — , Jun 14 '16 at 18:46
@ritwiksinha I don't think your comment does any help to me. — Harsh Sharma, Jun 14 '16 at 18:47
If you want to really help yourself then ask that part of a question that is troubling you, not the whole question. — , Jun 14 '16 at 18:48
@ritwiksinha will surely take care of that from ahead. Thanks for suggestion. — Harsh Sharma, Jun 14 '16 at 18:51

score 4 · Answer 1 · answered Jun 14 '16 at 18:45

4

We have $4R=\frac{abc}{\Delta}$ and $\sin A=\frac{2\Delta}{bc}$, hence the LHS equals $$ \frac{abc}{\Delta}\cdot\frac{8\Delta^3}{a^2 b^2 c^2}= \frac{8\Delta^2}{abc}.$$ Now exploit $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$ and Heron's formula.

answered Jun 14 '16 at 18:45

Jack D'Aurizio

353,855

score 2 · Accepted Answer · answered Jun 14 '16 at 18:48

2

By writing $a=2R \times \sin A$, $b=2R \times \sin B$ and $c=2R \times \sin C$, the relation is equivalent to $\sin A \sin B \sin C = \sin 2A + \sin 2B+\sin 2C$. By transforming the sum into product and writing $C= \pi - (A+B)$, you get the deserved equality.

answered Jun 14 '16 at 18:48

I. Stefan

384

See http://math.stackexchange.com/questions/154505/prove-that-sin2a-sin2b-sin2c-4-sina-sinb-sinc-when-a-b-c-are – lab bhattacharjee Jun 15 '16 at 04:56

Trigonometry-circumcircle and sides of triangle.

2 Answers2