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In a triangle ABC, prove that $a\cos(B-C)+b\cos(C-A)+c\cos(A-B)$ is equal to $\frac{abc}{R^2}$, where $a$, $b$, and $c$ are sides of the triangle and $R$ is the circumradius.

My work:-

By expanding Cosines equation becomes,

$a(\cos B \cos C+\sin B \sin C)+b(\cos C \cos A+\sin C \sin A)+c(\cos A \cos B+\sin A \sin B)$

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

$a(\cos B \cos C+\frac{bc}{4R^2})+b(\cos C \cos A+\frac{ca}{4R^2})+c(\cos A \cos B+\frac{ab}{4R^2})$

after saturation,

$\frac{3abc}{4R^2}+(a\cos B \cos C+b\cos C \cos A+c\cos A \cos B)$

I have no idea what to do next. There is no relation between Cos, abc and R to convert that term in bracket to abc and R.

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Dheeraj Kumar
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2 Answers2

2

$a\cos B\cos C+\cos A\left(b\cos C+c\cos B\right)=a\cos B\cos C+a\cos A$

$2R\sin A\cos B\cos C+2R\sin A\cos A\\=2R\sin A\cos B\cos C-2R\sin A\cos (B+C)\\=2R\sin A\sin B\sin C\\=\dfrac{abc}{4R^2}$

$\therefore\dfrac{3abc}{4R^2}+\dfrac{abc}{4R^2}=\dfrac{abc}{R^2}$

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As $a=2R\sin A$

and $\sin A=\sin[\pi-(B+C)]=\sin(B+C)$

$\implies a\cos(B-C)=2R\sin(B+C)\cos(B-C)=R[\sin2B+\sin2C]$

$\implies\sum a\cos(B-C)=2R[\sum\sin2A]$

Now use Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle

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lab bhattacharjee
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