If A + B + C = frac{3pi}{2}, then cos 2A + co | vedclass

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(D) Given $A + B + C = \frac{3\pi}{2}$.We need to evaluate $\cos 2A + \cos 2B + \cos 2C$.Using the formula $\cos X + \cos Y = 2\cos\frac{X+Y}{2}\cos\f

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$1 - 4\sin A \sin B \sin C$

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(D) Given $A + B + C = \frac{3\pi}{2}$.We need to evaluate $\cos 2A + \cos 2B + \cos 2C$.Using the formula $\cos X + \cos Y = 2\cos\frac{X+Y}{2}\cos\frac{X-Y}{2}$,we get:$\cos 2A + \cos 2B = 2\cos(A+B)\cos(A-B)$.Since $A+B = \frac{3\pi}{2} - C$,then $\cos(A+B) = \cos(\frac{3\pi}{2} - C) = -\sin C$.So,$\cos 2A + \cos 2B = -2\sin C \cos(A-B)$.Now,$\cos 2C = 1 - 2\sin^2 C$.Substituting these into the expression:$-2\sin C \cos(A-B) + 1 - 2\sin^2 C = 1 - 2\sin C(\cos(A-B) + \sin C)$.Since $\sin C = \sin(\frac{3\pi}{2} - (A+B)) = -\cos(A+B)$,$= 1 - 2\sin C(\cos(A-B) - \cos(A+B))$.Using $\cos(A-B) - \cos(A+B) = 2\sin A \sin B$,we get:$= 1 - 2\sin C(2\sin A \sin B) = 1 - 4\sin A \sin B \sin C$.

If $A + B + C = \frac{3\pi}{2},$ then $\cos 2A + \cos 2B + \cos 2C = $

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