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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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Verified Answer
The correct answer is:
$\sin \left(\frac{\pi}{4}-A\right)+\sin \left(\frac{\pi}{4}-B\right)+\sin \left(\frac{\pi}{4}-C\right)$
$$
\begin{aligned}
& \text { } 2\left(2 \cos \frac{A}{2} \cos \frac{B}{2}\right) \cos \frac{C}{2}-\cos \frac{\pi}{8} \\
& =2\left[\cos \left(\frac{A}{2}+\frac{B}{2}\right)+\cos \left(\frac{A}{2}-\frac{B}{2}\right)\right] \cos \frac{C}{2} \\
& =2\left[\cos \left(\frac{\pi}{8}-\frac{C}{2}\right)+\cos \left(\frac{A}{2}-\frac{B}{2}\right)\right] \cos \frac{C}{2}-\cos \frac{\pi}{8} \\
& =2 \cos \left(\frac{\pi}{8}-\frac{C}{2}\right) \cos \frac{C}{2}+2 \cos \left(\frac{A}{2}-\frac{B}{2}\right) \cos \frac{C}{2}-\cos \frac{\pi}{8}
\end{aligned}
$$
$$
\begin{aligned}
& =\cos \left(\frac{\pi}{8}\right)+\cos \left(\frac{\pi}{8}-C\right)+\cos \left(\frac{A}{2}-\frac{B}{2}+\frac{C}{2}\right) \\
& +\cos \left(\frac{A}{2}-\frac{B}{2}-\frac{C}{2}\right)-\cos \frac{\pi}{8} \\
& =\cos \left(\frac{\pi}{8}-C\right)+\cos \left(\frac{\pi}{8}-\frac{B}{2}-\frac{B}{2}\right)+\cos \left[\frac{A}{2}-\left(\frac{\pi}{8}-\frac{A}{2}\right)\right] \\
& =\cos \left(\frac{\pi}{8}-C\right)+\cos \left(\frac{\pi}{8}-B\right)+\cos \left(\frac{\pi}{8}-A\right)
\end{aligned}
$$
\begin{aligned}
& \text { } 2\left(2 \cos \frac{A}{2} \cos \frac{B}{2}\right) \cos \frac{C}{2}-\cos \frac{\pi}{8} \\
& =2\left[\cos \left(\frac{A}{2}+\frac{B}{2}\right)+\cos \left(\frac{A}{2}-\frac{B}{2}\right)\right] \cos \frac{C}{2} \\
& =2\left[\cos \left(\frac{\pi}{8}-\frac{C}{2}\right)+\cos \left(\frac{A}{2}-\frac{B}{2}\right)\right] \cos \frac{C}{2}-\cos \frac{\pi}{8} \\
& =2 \cos \left(\frac{\pi}{8}-\frac{C}{2}\right) \cos \frac{C}{2}+2 \cos \left(\frac{A}{2}-\frac{B}{2}\right) \cos \frac{C}{2}-\cos \frac{\pi}{8}
\end{aligned}
$$
$$
\begin{aligned}
& =\cos \left(\frac{\pi}{8}\right)+\cos \left(\frac{\pi}{8}-C\right)+\cos \left(\frac{A}{2}-\frac{B}{2}+\frac{C}{2}\right) \\
& +\cos \left(\frac{A}{2}-\frac{B}{2}-\frac{C}{2}\right)-\cos \frac{\pi}{8} \\
& =\cos \left(\frac{\pi}{8}-C\right)+\cos \left(\frac{\pi}{8}-\frac{B}{2}-\frac{B}{2}\right)+\cos \left[\frac{A}{2}-\left(\frac{\pi}{8}-\frac{A}{2}\right)\right] \\
& =\cos \left(\frac{\pi}{8}-C\right)+\cos \left(\frac{\pi}{8}-B\right)+\cos \left(\frac{\pi}{8}-A\right)
\end{aligned}
$$
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