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

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(A) Given $\cos 2 A+\cos 2 B+\cos 2 C$. Using the formula $\cos C+\cos D=2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$,we get: $=

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

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(A) Given $\cos 2 A+\cos 2 B+\cos 2 C$. Using the formula $\cos C+\cos D=2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$,we get: $=2 \cos (A+B) \cos (A-B)+\cos 2 C$. Since $A+B+C=\frac{3 \pi}{2}$,we have $A+B=\frac{3 \pi}{2}-C$. $=2 \cos \left(\frac{3 \pi}{2}-C\right) \cos (A-B)+\left(1-2 \sin ^2 C\right)$. $=2(-\sin C) \cos (A-B)+1-2 \sin ^2 C$. $=1-2 \sin C [\cos (A-B)+\sin C]$. Since $\sin C = \sin \left(\frac{3 \pi}{2}-(A+B)\right) = -\cos (A+B)$,we have: $=1-2 \sin C [\cos (A-B)-\cos (A+B)]$. 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 $\cos 2 A+\cos 2 B+\cos 2 C=$

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