In △ABC,1+4sin⁡π−A−4sin⁡π−B4sin⁡π−C4=

=1+2sin⁡π−A42sin⁡π−B4sin⁡π−C4=1+2sin⁡π−A4cos⁡π−B−π+C4−cos⁡π−B+π−C4=1+2sin⁡π−A4cos⁡B−C4−cos⁡π2−B+C4=1+2sin⁡π−A4cos⁡B−C4−2sin⁡π−A4sin⁡B+C4=1+sin⁡π−A+B−C4+sin⁡π−A−B+C4−cos⁡π−A−B−C4+cos⁡π−A+B+C4=1+sin⁡A+B+C−A+B−C4+sin⁡A+B+C−A−B+C4−1+cos⁡π−A+π−A4=sin⁡B2+sin⁡C2+sin⁡A2.

In △ABC,1+4sin⁡π−A−4sin⁡π−B4sin⁡π−C4=

=1+2sin⁡π−A42sin⁡π−B4sin⁡π−C4=1+2sin⁡π−A4cos⁡π−B−π+C4−cos⁡π−B+π−C4=1+2sin⁡π−A4cos⁡B−C4−cos⁡π2−B+C4=1+2sin⁡π−A4cos⁡B−C4−2sin⁡π−A4sin⁡B+C4=1+sin⁡π−A+B−C4+sin⁡π−A−B+C4−cos⁡π−A−B−C4+cos⁡π−A+B+C4=1+sin⁡A+B+C−A+B−C4+sin⁡A+B+C−A−B+C4−1+cos⁡π−A+π−A4=sin⁡B2+sin⁡C2+sin⁡A2.

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$In △ ABC, 1 + 4 \sin (\frac{π - A}{- 4}) \sin (\frac{π - B}{4}) \sin (\frac{π - C}{4}) =$

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

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

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

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

answer is A.

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Detailed Solution

$\begin{array}{l} = 1 + 2 \sin (\frac{π - A}{4}) [2 \sin \frac{π - B}{4} \sin \frac{π - C}{4}] \\ = 1 + 2 \sin \frac{π - A}{4} [\cos \frac{π - B - π + C}{4} - \cos \frac{π - B + π - C}{4}] \\ = 1 + 2 \sin (\frac{π - A}{4}) [\cos \frac{B - C}{4} - \cos (\frac{π}{2} - \frac{B + C}{4})] \\ = 1 + 2 \sin (\frac{π - A}{4}) \cos (\frac{B - C}{4}) - 2 \sin (\frac{π - A}{4}) \sin (\frac{B + C}{4}) \\ = 1 + \sin (\frac{π - A + B - C}{4}) + \sin (\frac{π - A - B + C}{4}) - \cos (\frac{π - A - B - C}{4}) + \cos (\frac{π - A + B + C}{4}) \\ = 1 + \sin (\frac{A + B + C - A + B - C}{4}) + \sin (\frac{A + B + C - A - B + C}{4}) - 1 + \cos (\frac{π - A + π - A}{4}) \\ = \sin \frac{B}{2} + \sin \frac{C}{2} + \sin \frac{A}{2} . \end{array}$