In △ABC,cos2⁡A2+cos2⁡B2+cos2⁡C2=

=1−sin2⁡A2+cos2⁡B2+cos2⁡C2=1+cos⁡A+B2cos⁡A−B2+1−sin2⁡C2=2+sin⁡C2cos⁡A−B2−cos⁡A+B2=2+2sin⁡A2sin⁡B2sin⁡C2

In △ABC,cos2⁡A2+cos2⁡B2+cos2⁡C2=

=1−sin2⁡A2+cos2⁡B2+cos2⁡C2=1+cos⁡A+B2cos⁡A−B2+1−sin2⁡C2=2+sin⁡C2cos⁡A−B2−cos⁡A+B2=2+2sin⁡A2sin⁡B2sin⁡C2

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$In △ ABC, \cos^{2} \frac{A}{2} + \cos^{2} \frac{B}{2} + \cos^{2} \frac{C}{2} =$

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

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

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

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

answer is D.

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Detailed Solution

$\begin{array}{l} = 1 - \sin^{2} \frac{A}{2} + \cos^{2} \frac{B}{2} + \cos^{2} \frac{C}{2} = 1 + \cos \frac{A + B}{2} \cos \frac{A - B}{2} + 1 - \sin^{2} \frac{C}{2} \\ = 2 + \sin \frac{C}{2} [\cos \frac{A - B}{2} - \cos \frac{A + B}{2}] = 2 + 2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \end{array}$