Question

In a $\Delta ABC$, $\sum a^2 \cos(B - C) =$

• A. $4abc$
• B. $3abc$
• C. $4a+b+c$
• D. $abc$

Hint:

For complex triangle identities involving summations, try substituting an equilateral triangle ($A=B=C=60^\circ$). If $a=b=c=1$, the expression becomes $1^2 \cos(0) + 1^2 \cos(0) + 1^2 \cos(0) = 3$. Check which option gives 3 when $a=b=c=1$.

Answer 1

The Correct Option is B

1. Expansion of the Summation: The expression $\sum a^2 \cos(B - C)$ means: $$a^2 \cos(B - C) + b^2 \cos(C - A) + c^2 \cos(A - B)$$

2. Simplify one term using Sine Rule: From Sine Rule, $a = 2R \sin A$. Since $A = 180^\circ - (B + C)$, we have $\sin A = \sin(B + C)$. $$a^2 \cos(B - C) = (2R \sin A) a \cos(B - C)$$ $$= 2R \sin(B + C) a \cos(B - C)$$ Using the identity $2 \sin X \cos Y = \sin(X+Y) + \sin(X-Y)$: $$a^2 \cos(B - C) = Ra [\sin(B+C+B-C) + \sin(B+C-B+C)]$$ $$= Ra [\sin 2B + \sin 2C]$$ $$= Ra [2 \sin B \cos B + 2 \sin C \cos C]$$

3. Completing the Sum: Repeating this for all terms and substituting $b = 2R \sin B$ and $c = 2R \sin C$: $$\sum a^2 \cos(B - C) = 3abc$$ This is a standard identity in the properties of triangles.