Question

In $△ABC,\sum a^3\cos (B-C)=$

• A. $abc$
• B. $2abc$
• C. $3abc$
• D. $0$

Answer 1

The correct option is C $3abc$

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

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

Using sine rule, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

we know that $A+B+C=\pi$

$\Rightarrow a=k\sin A=k\sin (\pi -(B+C))=k\sin (B+C)$

$\Rightarrow b=k\sin B=k\sin (\pi -(C+A))=k\sin (C+A)$

$\Rightarrow c=k\sin C=k\sin (\pi -(A+B))=k\sin (A+B)$

L.H.S$=a^{2}k\sin (B+C)\cos (B-C)+b^{2}k\sin (C+A)\cos (C-A)+c^{2}k\sin (A+B)\cos (A-B)$

$=\frac{k}{2}[a^{2}2\sin (B+C)\cos (B-C)+b^{2}2\sin (C+A)\cos (C-A)+c^{2}2\sin (A+B)\cos (A-B)]$

$=\frac{k}{2}[a^2(\sin 2B+\sin 2C)+b^2(\sin 2C+\sin 2A)+c^2(\sin 2A+\sin 2B)]$

$=\frac{k}{2}[a^2(2\sin B\cos B+2\sin C\cos C)+b^2(2\sin C\cos C+2\sin A\cos A)+c^2(2\sin A\cos A+2\sin B\cos B)]$

$=a^2(k\sin B\cos B+k\sin C\cos C)+b^2(k\sin C\cos C+k\sin A\cos A)+c^2(k\sin A\cos A+k\sin B\cos B)$

We know that $a=k\sin A,b=k\sin B,c=k\sin C$ we have

$=a^2(b\cos B+c\cos C)+b^2(c\cos C+a\cos A)+c^2(a\cos A+b\cos B)$

$=ab(a\cos B+b\cos A)+bc(b\cos C+c\cos B)+ca(c\cos A+a\cos C)$

Using projection rule, we have $c=a\cos B+b\cos A,a=b\cos C+c\cos B,b=c\cos A+a\cos C$

$=abc+bca+cab=3abc$