In any triangle $A B C, a^{3} cos (B - C) + b^{3} cos (C - A) + c^{3} cos (A - B)$ is equal ?

6 a b c

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9 a b c

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3 a b c

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N o n e

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Solution

The correct option is A $3 a b c$
We know,
Law of sines, $k = \frac{s i n A}{a} = \frac{s i n B}{b} = \frac{s i n C}{c}$
$s i n 2 x + s i n 2 y = 2 s i n (x + y) c o s (x - y)$
$c o s (x - y) = \frac{s i n 2 x + s i n 2 y}{2 s i n (x + y)}$
= $\frac{s i n x c o s x + s i n y c o s y}{s i n (x + y)}$
Therefore,
$a^{3} c o s (B - C) = a^{3} \frac{s i n B c o s B + s i n C c o s C}{s i n (B + C)}$ = $a^{3} \frac{b k c o s B + c k c o s C}{s i n (180^{0} - A)}$
$= a^{3} \frac{b k c o s B + c k c o s C}{a k}$
Therefore,
$a^{3} c o s (B - C) + b^{3} c o s (C - A) + c^{3} c o s (A - B)$ = $a^{2} (b c o s B + c c o s C) + b^{2} (c c o s C + a c o s A) + c^{2} (a c o s A + b c o s B)$
= $a^{2} b c o s B + a b^{2} c o s A + a^{2} c c o s C + c^{2} a c o s A + b^{2} c c o s C + c^{2} b c o s B$
= $a b (a c o s B + b c o s A) + a c (a c o s C + c c o s A) + b c (b c o s C + c c o s B)$
Since, $a = b c o s C + c c o s B, b = c c o s A + a c o s C, c = a c o s B + b c o s A$
So,
$a^{3} c o s (B - C) + b^{3} c o s (C - A) + c^{3} c o s (A - B)$ = $a^{2} (b c o s B + c c o s C) + b^{2} (c c o s C + a c o s A) + c^{2} (a c o s A + b c o s B)$ = $a b c + a b c + a b c$ = $3 a b c$

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