If A+B+C=270^{circ},then cos 2A+cos 2B+cos 2C | vedclass

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(C) Given $A+B+C=270^{\circ}$,so $A+B = 270^{\circ}-C$. Using the identity $\cos 2A + \cos 2B = 2 \cos(A+B) \cos(A-B)$: $\cos 2A + \cos 2B + \cos 2C =

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

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(C) Given $A+B+C=270^{\circ}$,so $A+B = 270^{\circ}-C$. Using the identity $\cos 2A + \cos 2B = 2 \cos(A+B) \cos(A-B)$: $\cos 2A + \cos 2B + \cos 2C = 2 \cos(A+B) \cos(A-B) + (2 \cos^2 C - 1)$ $= 2 \cos(270^{\circ}-C) \cos(A-B) + 2 \cos^2 C - 1$ $= 2(-\sin C) \cos(A-B) + 2 \cos^2 C - 1$ $= -2 \sin C \cos(A-B) + 2(1-\sin^2 C) - 1$ $= 1 - 2 \sin C [\cos(A-B) + \sin C]$ Since $\sin C = \sin(270^{\circ}-(A+B)) = -\cos(A+B)$: $= 1 - 2 \sin C [\cos(A-B) - \cos(A+B)]$ $= 1 - 2 \sin C [2 \sin A \sin B]$ $= 1 - 4 \sin A \sin B \sin C$.

If $A+B+C=270^{\circ}$,then $\cos 2A+\cos 2B+\cos 2C$ is equal to:

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