If A+B+C=180∘ then 4cos⁡π−A4cos⁡π−B4cos⁡π−C4=

=2cos⁡π−A42cos⁡π−B4cos⁡π−C4=2cos⁡π−A4cos⁡π−B+π−C4+cos⁡π−B−π+C4=2cos⁡π−A4cos⁡π2−B+C4+cos⁡B−C4=2cos⁡π−A4sin⁡B+C4+2cos⁡π−A4cos⁡B−C4=sin⁡π−A+B+C4−sin⁡π−A−B−C4+cos⁡π−A+B−C4+cos⁡π−A−B+C4=sin⁡π−A+π−A4−0+cos⁡A+B+C−A+B−C4+cos⁡A+B+C−A−B+C4=cos⁡A2+cos⁡B2+cos⁡C2

If A+B+C=180∘ then 4cos⁡π−A4cos⁡π−B4cos⁡π−C4=

=2cos⁡π−A42cos⁡π−B4cos⁡π−C4=2cos⁡π−A4cos⁡π−B+π−C4+cos⁡π−B−π+C4=2cos⁡π−A4cos⁡π2−B+C4+cos⁡B−C4=2cos⁡π−A4sin⁡B+C4+2cos⁡π−A4cos⁡B−C4=sin⁡π−A+B+C4−sin⁡π−A−B−C4+cos⁡π−A+B−C4+cos⁡π−A−B+C4=sin⁡π−A+π−A4−0+cos⁡A+B+C−A+B−C4+cos⁡A+B+C−A−B+C4=cos⁡A2+cos⁡B2+cos⁡C2

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$If A + B + C = 180^{\circ} then 4 \cos (\frac{π - A}{4}) \cos (\frac{π - B}{4}) \cos (\frac{π - C}{4}) =$

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cos A + cos B + cos c

$\cos \frac{A}{2} + \cos \frac{B}{2} + \cos \frac{C}{2}$

sin A + sin B + sin C

$\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} = 2 \cos \frac{π - A}{4} 2 \cos \frac{π - B}{4} \cos \frac{π - C}{4} \\ = 2 \cos \frac{π - A}{4} [\cos \frac{π - B + π - C}{4} + \cos \frac{π - B - π + C}{4}] \\ = 2 \cos \frac{π - A}{4} (\cos (\frac{π}{2} - \frac{B + C}{4}) + \cos \frac{B - C}{4}) = 2 \cos \frac{π - A}{4} \sin \frac{B + C}{4} + 2 \cos \frac{π - A}{4} \cos \frac{B - C}{4} \end{array}$
$\begin{array}{l} = \sin (\frac{π - A + B + C}{4}) - \sin (\frac{π - A - B - C}{4}) + \cos \frac{π - A + B - C}{4} + \cos \frac{π - A - B + C}{4} \\ = \sin (\frac{π - A + π - A}{4}) - 0 + \cos (\frac{A + B + C - A + B - C}{4}) + \cos (\frac{A + B + C - A - B + C}{4}) \\ = \cos \frac{A}{2} + \cos \frac{B}{2} + \cos \frac{C}{2} \end{array}$