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: $\cos \left(\frac{B+2C+3A}{2}ight) + \cos \left(\frac{A-B}{2}ight)$ Re

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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 first term using $A+B+C = \pi$: $\frac{B+2C+3A}{2} = \frac{(A+B+C) + C + 2A}{2} = \frac{\pi + C + 2A}{2} = \frac{\pi}{2} + \frac{2A+C}{2}$ Alternatively,rewrite it as: $\frac{B+2C+3A}{2} = \frac{2(A+B+C) + A-B}{2} = \frac{2\pi + (A-B)}{2} = \pi + \frac{A-B}{2}$ Substituting this back into the expression: $\cos \left(\pi + \frac{A-B}{2}ight) + \cos \left(\frac{A-B}{2}ight)$ 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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