If cos A+cos B+cos C=0 and sin A+sin B+sin C= | vedclass

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(D) Given $\cos A+\cos B+\cos C=0$ and $\sin A+\sin B+\sin C=0$. Let $x_1 = \cos A + i \sin A$,$x_2 = \cos B + i \sin B$,and $x_3 = \cos C + i \sin C$

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(D) Given $\cos A+\cos B+\cos C=0$ and $\sin A+\sin B+\sin C=0$. Let $x_1 = \cos A + i \sin A$,$x_2 = \cos B + i \sin B$,and $x_3 = \cos C + i \sin C$. Then $x_1 + x_2 + x_3 = (\cos A + \cos B + \cos C) + i(\sin A + \sin B + \sin C) = 0 + 0i = 0$. Since $|x_1| = |x_2| = |x_3| = 1$,we have $x_1 + x_2 + x_3 = 0$. This implies $x_1^2 + x_2^2 + x_3^2 + 2(x_1x_2 + x_2x_3 + x_3x_1) = 0$. Also,$x_1x_2x_3(\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3}) = 0$,which means $\bar{x_1} + \bar{x_2} + \bar{x_3} = 0$. Using the property of complex numbers on the unit circle,$x_1+x_2+x_3=0$ implies $x_1x_2+x_2x_3+x_3x_1=0$. Thus,$x_1, x_2, x_3$ are roots of $z^3 - k = 0$ where $k = x_1x_2x_3$. Then $x_1/x_2 + x_2/x_1 = 2 \cos(A-B)$. From $x_1+x_2 = -x_3$,squaring gives $x_1^2 + x_2^2 + 2x_1x_2 = x_3^2$. Dividing by $x_1x_2$,we get $x_1/x_2 + x_2/x_1 + 2 = x_3^2/(x_1x_2) = x_3^3/(x_1x_2x_3) = x_3^3/k$. Since $x_3^3 = k$,we have $2 \cos(A-B) + 2 = 1$,so $2 \cos(A-B) = -1$. Therefore,$\cos(A-B) = -\frac{1}{2}$.

List-$I$	List-$II$
$(i)$ $a \bar{a}$	$(A)$ $-\frac{\pi}{3}$
$(ii)$ $\arg \left(\frac{1}{\bar{a}}\right)$	$(B)$ $-i \sqrt{3}$
$(iii)$ $a - \bar{a}$	$(C)$ $2i / \sqrt{3}$
$(iv)$ $\operatorname{Im}\left(\frac{4}{3a}\right)$	$(D)$ $1$
	$(E)$ $\pi / 3$
	$(F)$ $\frac{2}{\sqrt{3}}$

If $\cos A+\cos B+\cos C=0$ and $\sin A+\sin B+\sin C=0$,then $\cos (A-B)=$

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