If A+B+C=180∘ then cos⁡3A+cos⁡3B+cos⁡3C=

=2cos⁡3A=3B2cos⁡3A−3B2+1−2sin2⁡3C2=1−2sin⁡3C2cos⁡3A−3B2−cos⁡3A+3B2=1−4sin⁡3A2sin⁡3B2sin⁡3C2

If A+B+C=180∘ then cos⁡3A+cos⁡3B+cos⁡3C=

=2cos⁡3A=3B2cos⁡3A−3B2+1−2sin2⁡3C2=1−2sin⁡3C2cos⁡3A−3B2−cos⁡3A+3B2=1−4sin⁡3A2sin⁡3B2sin⁡3C2

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$If A + B + C = 180^{\circ} then \cos 3 A + \cos 3 B + \cos 3 C =$

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$4 \cos \frac{3 A}{2} \cos \frac{3 B}{2} \cos \frac{3 C}{2}$

$- 4 \cos \frac{3 A}{2} \cos \frac{3 B}{2} \cos \frac{3 C}{2}$

$1 - 4 \cos \frac{3 A}{2} \cos \frac{3 B}{2} \cos \frac{3 C}{2}$

$1 - 4 \sin \frac{3 A}{2} \sin \frac{3 B}{2} \sin \frac{3 C}{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} = 2 \cos \frac{3 A = 3 B}{2} \cos \frac{3 A - 3 B}{2} + 1 - 2 \sin^{2} \frac{3 C}{2} \\ = 1 - 2 \sin \frac{3 C}{2} [\cos (\frac{3 A - 3 B}{2}) - \cos (\frac{3 A + 3 B}{2})] = 1 - 4 \sin \frac{3 A}{2} \sin \frac{3 B}{2} \sin \frac{3 C}{2} \end{array}$