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

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(D) Given equations are: $\cos A + \cos B + \cos C = 0$  $(i)$ $\sin A + \sin B + \sin C = 0$  $(ii)$ Squaring both equations $(i)$ and $(ii)$ and add

Vedclass · Accepted Answer

$-\frac{3}{2}$

Vedclass · Answer

(D) Given equations are: $\cos A + \cos B + \cos C = 0$  $(i)$ $\sin A + \sin B + \sin C = 0$  $(ii)$ Squaring both equations $(i)$ and $(ii)$ and adding them: $(\cos A + \cos B + \cos C)^2 + (\sin A + \sin B + \sin C)^2 = 0^2 + 0^2$ $(\cos^2 A + \cos^2 B + \cos^2 C + 2\cos A \cos B + 2\cos B \cos C + 2\cos C \cos A) + (\sin^2 A + \sin^2 B + \sin^2 C + 2\sin A \sin B + 2\sin B \sin C + 2\sin C \sin A) = 0$ Grouping the terms: $(\cos^2 A + \sin^2 A) + (\cos^2 B + \sin^2 B) + (\cos^2 C + \sin^2 C) + 2(\cos A \cos B + \sin A \sin B) + 2(\cos B \cos C + \sin B \sin C) + 2(\cos C \cos A + \sin C \sin A) = 0$ Using the identity $\sin^2 	heta + \cos^2 	heta = 1$ and $\cos(x-y) = \cos x \cos y + \sin x \sin y$: $1 + 1 + 1 + 2[\cos(A-B) + \cos(B-C) + \cos(C-A)] = 0$ $3 + 2[\cos(A-B) + \cos(B-C) + \cos(C-A)] = 0$ $2[\cos(A-B) + \cos(B-C) + \cos(C-A)] = -3$ $\cos(A-B) + \cos(B-C) + \cos(C-A) = -\frac{3}{2}$

If $\cos A + \cos B + \cos C = 0$ and $\sin A + \sin B + \sin C = 0$,then find the value of $\cos (A - B) + \cos (B - C) + \cos (C - A)$.

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