If A+B+C=4S,then cos (2S-A)+cos (2S-B)-cos (2 | vedclass

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(D) Given $A+B+C=4S$. Now,consider the expression $\cos (2S-A)+\cos (2S-B)-\cos (2S-C)-\cos 2S$. $= [\cos (2S-A)+\cos (2S-B)] - [\cos (2S-C)+\cos 2S]$

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$4 \sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$

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(D) Given $A+B+C=4S$. Now,consider the expression $\cos (2S-A)+\cos (2S-B)-\cos (2S-C)-\cos 2S$. $= [\cos (2S-A)+\cos (2S-B)] - [\cos (2S-C)+\cos 2S]$. Using the formula $\cos 	heta + \cos \phi = 2 \cos \left(\frac{	heta+\phi}{2}ight) \cos \left(\frac{	heta-\phi}{2}ight)$,we get: $= 2 \cos \left(\frac{4S-A-B}{2}ight) \cos \left(\frac{B-A}{2}ight) - 2 \cos \left(\frac{4S-C}{2}ight) \cos \left(\frac{-C}{2}ight)$. Since $A+B+C=4S$,we have $4S-A-B=C$ and $4S-C=A+B$. $= 2 \cos \left(\frac{C}{2}ight) \cos \left(\frac{B-A}{2}ight) - 2 \cos \left(\frac{A+B}{2}ight) \cos \left(\frac{C}{2}ight)$. $= 2 \cos \left(\frac{C}{2}ight) \left[ \cos \left(\frac{B-A}{2}ight) - \cos \left(\frac{A+B}{2}ight) ight]$. Using $\cos 	heta - \cos \phi = 2 \sin \left(\frac{	heta+\phi}{2}ight) \sin \left(\frac{\phi-	heta}{2}ight)$,we get: $= 2 \cos \left(\frac{C}{2}ight) \left[ 2 \sin \left(\frac{B}{2}ight) \sin \left(\frac{A}{2}ight) ight]$. $= 4 \sin \left(\frac{A}{2}ight) \sin \left(\frac{B}{2}ight) \cos \left(\frac{C}{2}ight)$.

If $A+B+C=4S$,then $\cos (2S-A)+\cos (2S-B)-\cos (2S-C)-\cos 2S=$

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