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

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(D) Given: $A + B + C = \frac{3\pi}{2} \implies A + B = \frac{3\pi}{2} - C$.Consider the expression: $\cos 2A + \cos 2B + \cos 2C$.Using the formula $

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

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(D) Given: $A + B + C = \frac{3\pi}{2} \implies A + B = \frac{3\pi}{2} - C$.Consider the expression: $\cos 2A + \cos 2B + \cos 2C$.Using the formula $\cos X + \cos Y = 2\cos\left(\frac{X+Y}{2}\right)\cos\left(\frac{X-Y}{2}\right)$:$\cos 2A + \cos 2B = 2\cos(A+B)\cos(A-B)$.Substitute $A+B = \frac{3\pi}{2} - C$:$2\cos\left(\frac{3\pi}{2} - C\right)\cos(A-B) = 2(-\sin C)\cos(A-B) = -2\sin C \cos(A-B)$.Now,$\cos 2C = 1 - 2\sin^2 C$.So,$\cos 2A + \cos 2B + \cos 2C = 1 - 2\sin C \cos(A-B) - 2\sin^2 C$.$= 1 - 2\sin C [\cos(A-B) + \sin C]$.Since $C = \frac{3\pi}{2} - (A+B)$,$\sin C = \sin\left(\frac{3\pi}{2} - (A+B)\right) = -\cos(A+B)$.$= 1 - 2\sin C [\cos(A-B) - \cos(A+B)]$.Using $\cos(A-B) - \cos(A+B) = 2\sin A \sin B$:$= 1 - 2\sin C [2\sin A \sin B] = 1 - 4\sin A \sin B \sin C$.

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

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