Giúp mình với ạ! Câu này không biết giải cách nào

Đáp án:

$D.$

Giải thích các bước giải:

$\sin 2x=2\sin x\cos x\\A=\sin\dfrac{\pi}{48}\cos\dfrac{\pi}{48}\cos\dfrac{\pi}{24}\cos\dfrac{\pi}{12}\cos\dfrac{\pi}{6}\\ =\dfrac{1}{2}.2\sin\dfrac{\pi}{48}\cos\dfrac{\pi}{48}\cos\dfrac{\pi}{24}\cos\dfrac{\pi}{12}\cos\dfrac{\pi}{6}\\ =\dfrac{1}{2}\sin\dfrac{\pi}{24}\cos\dfrac{\pi}{24}\cos\dfrac{\pi}{12}\cos\dfrac{\pi}{6}\\ =\dfrac{1}{4}.2\sin\dfrac{\pi}{24}\cos\dfrac{\pi}{24}\cos\dfrac{\pi}{12}\cos\dfrac{\pi}{6}\\ =\dfrac{1}{4}\sin\dfrac{\pi}{12}\cos\dfrac{\pi}{12}\cos\dfrac{\pi}{6}\\ =\dfrac{1}{8}.2\sin\dfrac{\pi}{12}\cos\dfrac{\pi}{12}\cos\dfrac{\pi}{6}\\ =\dfrac{1}{8}\sin\dfrac{\pi}{6}\cos\dfrac{\pi}{6}\\ =\dfrac{1}{16}.2\sin\dfrac{\pi}{6}\cos\dfrac{\pi}{6}\\ =\dfrac{1}{16}\sin\dfrac{\pi}{3}\\ =\dfrac{\sqrt{3}}{32}.$