Question under solution of triangles

Question

Prove that:

$a^3\cos(B-C) + b^3\cos(C-A) + c^3\cos(A-B) = 3abc$

Here a,b,c are the sides of the triangle ABC opposite to angles A, B, C respectively.

My attempt: I tried putting $a = 2R \sin(A)$ and replacing $\sin(B+C) = \sin(A)$ but that didn't help out. I tried applying AM $\ge$ GM but that too did not work out for this question.

Please help me out!!!

https://math.stackexchange.com/questions/2861484/how-to-prove-a3-cos-b-c-b3-cos-c-a-c3-cos-a-b-3abc — lab bhattacharjee, Jun 25 '19 at 05:31

score 1 · Answer 1 · answered Jun 25 '19 at 05:23

1

$$\sum a^3\cos(B-C)$$

$$=\sum 2Ra^2\sin(A)\cos(B-C)$$

$$=\sum Ra^2(\sin(A+B-C)+\sin(A-B+C))$$

$$=\sum Ra^2(\sin(2C)+\sin(2B))$$

$$=\sum a^2 c\cos C + \sum a^2b\cos B$$

$$=\sum a^2c\cos C + \sum c^2a\cos A$$

$$=\sum ac(a\cos C+c\cos A)$$

$$=\sum abc=\fbox{3abc}$$

answered Jun 25 '19 at 05:23

Amit Rajaraman

3,248

Thankyou for helping.. – Prajwal Turkar Jun 25 '19 at 05:25

Question under solution of triangles

1 Answers1