In a triangle ABC,cos left(frac{B+2C+3A}{2}ri | vedclass

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(B) In $	riangle ABC$,we know that $A+B+C = \pi$.Consider the expression:$\cos \left(\frac{B+2C+3A}{2}ight) + \cos \left(\frac{A-B}{2}ight)$Rewri

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(B) In $	riangle ABC$,we know that $A+B+C = \pi$.Consider the expression:$\cos \left(\frac{B+2C+3A}{2}ight) + \cos \left(\frac{A-B}{2}ight)$Rewrite the argument of the first cosine term using $A+B+C = \pi$:$\frac{B+2C+3A}{2} = \frac{(A+B+C) + A + C}{2} = \frac{\pi + A + C}{2} = \frac{\pi}{2} + \frac{A+C}{2}$Wait,let us re-evaluate the expression:$\frac{B+2C+3A}{2} = \frac{2A+2B+2C + A-B}{2} = \frac{2\pi + (A-B)}{2} = \pi + \frac{A-B}{2}$Substituting this back:$\cos \left(\pi + \frac{A-B}{2}ight) + \cos \left(\frac{A-B}{2}ight)$Using the identity $\cos(\pi + 	heta) = -\cos(	heta)$:$-\cos \left(\frac{A-B}{2}ight) + \cos \left(\frac{A-B}{2}ight) = 0$

In a $\triangle ABC$,$\cos \left(\frac{B+2C+3A}{2}\right) + \cos \left(\frac{A-B}{2}\right)$ is equal to

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