Đặt $A=sin 1^o.sin 2^o.sin3^o\dots.sin88^o.sin 89^o.sin90^o$
Ta có: $\sin x\cdot\sin\left ( \dfrac{\pi}{3}-x \right )\sin\left ( \dfrac{\pi}{3}+x \right )=\dfrac{1}{4}\sin3x$
Suy ra:
$A=\left (\sin 1^o\sin59^o\sin 61^o \right )\left ( \sin 2^o\sin58^{\circ}\sin 62^{\circ} \right )\dots\left (\sin 29^{\circ}\sin31^{\circ}\sin 89^{\circ} \right )\sin30^{\circ}\sin60^{\circ}$
$=\left (\dfrac{1}{4} \right )^{29}\dfrac{\sqrt3}{4}\sin3^{\circ}\sin6^{\circ}\dots\sin87^{\circ}$
$=\left (\dfrac{1}{4} \right )^{30}\sqrt3\left (\sin3^{\circ} \sin57^{\circ}\sin63^{\circ} \right )\left ( \sin6^{\circ}\sin54^{\circ}\sin66^{\circ} \right )\dots\left ( \sin27^{\circ}\sin33^{\circ}\sin87^{\circ} \right )\sin30^{\circ}\sin60^{\circ}$
$=\left (\dfrac{1}{4} \right )^{40}3\sin9^{\circ}\sin18^{\circ}\dots\sin81^{\circ}$
$=\left (\dfrac{1}{4} \right )^{40}3\left (\sin9^{\circ}\cos9^{\circ} \right )\left (\sin18^{\circ}\cos18^{\circ} \right )\left ( \sin27^{\circ}\cos27^{\circ} \right )\left ( \sin36^{\circ}\cos36^{\circ} \right )\sin45^{\circ}$
$=\left (\dfrac{1}{4} \right )^{43}6\sqrt2\sin18^{\circ}\sin36^{\circ}\sin54^{\circ}\sin72^{\circ}$
$=\left (\dfrac{1}{4} \right )^{43}3\sqrt2\left (\sin^236^{\circ}\cdot\cos36^{\circ} \right )$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
