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  4. If A+B+C=(π/4), then 4 cos (A/2) cos (B/2) cos (C/2)- cos (π/8)=

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

TS EAMCET 2018

Solution:

$2\left(2 \cos \frac{A}{2} \cos \frac{B}{2}\right) \cos \frac{c}{2}-\cos \frac{\pi}{8}$
$=2\left[\cos \left(\frac{A}{2}+\frac{B}{2}\right)+\cos \left(\frac{A}{2}-\frac{B}{2}\right)\right] \cos \frac{c}{2}$
$=2\left[\cos \left(\frac{\pi}{8}-\frac{C}{2}\right)+\cos \left(\frac{A}{2}-\frac{B}{2}\right)\right] \cos \frac{C}{2}-\cos \frac{\pi}{8}$
$=2 \cos \left(\frac{\pi}{8}-\frac{C}{2}\right) \cos \frac{C}{2}+2 \cos \left(\frac{A}{2}-\frac{B}{2}\right) \cos \frac{C}{2}-\cos \frac{\pi}{8}$
$=\cos \left(\frac{\pi}{8}\right)+\cos \left(\frac{\pi}{8}-C\right)+\cos \left(\frac{A}{2}-\frac{B}{2}+\frac{C}{2}\right)$
$+\cos \left(\frac{A}{2}-\frac{B}{2}-\frac{C}{2}\right)-\cos \frac{\pi}{8}$
$=\cos \left(\frac{\pi}{8}-c\right)+\cos \left(\frac{\pi}{8}-\frac{B}{2}-\frac{B}{2}\right)+\cos \left[\frac{A}{2}-\left(\frac{\pi}{8}-\frac{A}{2}\right)\right]$
$=\cos \left(\frac{\pi}{8}-C\right)+\cos \left(\frac{\pi}{8}-B\right)+\cos \left(\frac{\pi}{8}-A\right)$