In triangle ABC,if cos A + cos C = 4 sin^2 fr | vedclass

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(B) Given $\cos A + \cos C = 4 \sin^2 \frac{B}{2}$.Using the sum-to-product formula,$2 \cos \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)

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$3: 2$

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(B) Given $\cos A + \cos C = 4 \sin^2 \frac{B}{2}$.Using the sum-to-product formula,$2 \cos \left(\frac{A+C}{2}ight) \cos \left(\frac{A-C}{2}ight) = 4 \sin^2 \frac{B}{2}$.Since $A+B+C = 180^{\circ}$,$\frac{A+C}{2} = 90^{\circ} - \frac{B}{2}$,so $\cos \left(\frac{A+C}{2}ight) = \sin \frac{B}{2}$.Substituting this,$2 \sin \frac{B}{2} \cos \left(\frac{A-C}{2}ight) = 4 \sin^2 \frac{B}{2}$.Dividing by $2 \sin \frac{B}{2}$ (as $\sin \frac{B}{2} 
eq 0$),we get $\cos \left(\frac{A-C}{2}ight) = 2 \sin \frac{B}{2}$.Using $\sin \frac{B}{2} = \cos \left(\frac{A+C}{2}ight)$,we have $\cos \left(\frac{A-C}{2}ight) = 2 \cos \left(\frac{A+C}{2}ight)$.Expanding,$\cos \frac{A}{2} \cos \frac{C}{2} + \sin \frac{A}{2} \sin \frac{C}{2} = 2 (\cos \frac{A}{2} \cos \frac{C}{2} - \sin \frac{A}{2} \sin \frac{C}{2})$.Rearranging gives $3 \sin \frac{A}{2} \sin \frac{C}{2} = \cos \frac{A}{2} \cos \frac{C}{2}$,or $	an \frac{A}{2} 	an \frac{C}{2} = \frac{1}{3}$.Using $	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$,this simplifies to $\frac{s-b}{s} = \frac{1}{3}$,which implies $3s - 3b = s$,so $2s = 3b$.Since $2s = a+b+c$,we have $a+b+c = 3b$,which means $a+c = 2b$.The ratio of the perimeter $(a+b+c)$ to $(a+c)$ is $\frac{3b}{2b} = \frac{3}{2}$ or $3: 2$.

In $\triangle ABC$,if $\cos A + \cos C = 4 \sin^2 \frac{B}{2}$,then the ratio between the perimeter of the triangle and $(a+c)$ is

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