Nilai dari sin 75 . sin 15 . cos 135 adalah...

-√2/8

Pembahasan

Untuk menyelesaikan soal ini, kita perlu menghitung nilai dari $\sin 75^\circ \cdot \sin 15^\circ \cdot \cos 135^\circ$. 
Kita tahu bahwa $\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$.
Juga, $\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$.
Dan $\cos 135^\circ = \cos(180^\circ - 45^\circ) = -\cos 45^\circ = -\frac{\sqrt{2}}{2}$.
Maka, $\sin 75^\circ \cdot \sin 15^\circ = (\frac{\sqrt{6} + \sqrt{2}}{4}) \cdot (\frac{\sqrt{6} - \sqrt{2}}{4}) = \frac{(\sqrt{6})^2 - (\sqrt{2})^2}{16} = \frac{6-2}{16} = \frac{4}{16} = \frac{1}{4}$.
Jadi, $\sin 75^\circ \cdot \sin 15^\circ \cdot \cos 135^\circ = \frac{1}{4} \cdot (-\frac{\sqrt{2}}{2}) = -\frac{\sqrt{2}}{8}$.