If A+B+C=frac{pi}{4},then 4 cos frac{A}{2} co | vedclass

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(B) Given $A+B+C=\frac{\pi}{4}$,we have $\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{8}$.Consider the expression $E = 4 \cos \frac{A}{2} \cos \frac

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$\cos \left(\frac{\pi}{8}-A\right)+\cos \left(\frac{\pi}{8}-B\right)+\cos \left(\frac{\pi}{8}-C\right)$

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(B) Given $A+B+C=\frac{\pi}{4}$,we have $\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{8}$.Consider the expression $E = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}-\cos \frac{\pi}{8}$.Using $2 \cos x \cos y = \cos(x+y) + \cos(x-y)$:$E = 2 \left[ \cos \left( \frac{A+B}{2} \right) + \cos \left( \frac{A-B}{2} \right) \right] \cos \frac{C}{2} - \cos \frac{\pi}{8}$.Since $\frac{A+B}{2} = \frac{\pi}{8} - \frac{C}{2}$,we have:$E = 2 \cos \left( \frac{\pi}{8} - \frac{C}{2} \right) \cos \frac{C}{2} + 2 \cos \left( \frac{A-B}{2} \right) \cos \frac{C}{2} - \cos \frac{\pi}{8}$.Using $2 \cos x \cos y = \cos(x+y) + \cos(x-y)$ again:$E = \left[ \cos \frac{\pi}{8} + \cos \left( \frac{\pi}{8} - C \right) \right] + \left[ \cos \left( \frac{A-B+C}{2} \right) + \cos \left( \frac{A-B-C}{2} \right) \right] - \cos \frac{\pi}{8}$.Since $\frac{A-B+C}{2} = \frac{A+B+C-2B}{2} = \frac{\pi}{8} - B$ and $\frac{A-B-C}{2} = \frac{A-(B+C)}{2} = \frac{A-(\frac{\pi}{4}-A)}{2} = A - \frac{\pi}{8}$,and $\cos(A-\frac{\pi}{8}) = \cos(\frac{\pi}{8}-A)$:$E = \cos \left( \frac{\pi}{8} - C \right) + \cos \left( \frac{\pi}{8} - B \right) + \cos \left( \frac{\pi}{8} - A \right)$.

If $A+B+C=\frac{\pi}{4}$,then $4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}-\cos \frac{\pi}{8}=$

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