<s>[INST] Find the value of $\sin25^° \sin35^° \sin85^°$ Trigo problem : 
Find the value of $\sin25^° \sin35^° \sin85^°$.
My approach : 
Using $2\sin A\sin B = \cos(A-B) -\cos(A+B)$
$$
\begin{align}
& \phantom{={}}[\cos10^{°} -\cos60^°] \sin85^° \\
& = \frac{1}{2}[2\cos10^°\sin85^° -2\cos60^° \sin85^°] \\
& = \frac{1}{2}[2\cos10^°\sin85^° -2 \frac{1}{2} \sin85^°] \\
& = \frac{1}{2}[2\cos10^°\sin85^° -\sin85^°] \\
& = \frac{1}{2} [\sin 95^° + \sin 75^° -\sin85^°]
\end{align}
$$
After this I am unable to solve further.... please guide... thanks.
 [/INST] \begin{align}
& \phantom{={}} \sin25^{°}\sin35^{°}\sin85^{°} \\
& =\sin85^{°}\sin35^{°}\sin25^{°} \\
& =\frac{1}{2}(2\sin85^{°}\sin35^{°})\sin25^{°} \\
& =\frac{1}{2}[ \cos50^{°} -\cos120^{°}] \sin25^{°}  \\
& = \frac{1}{4}[ 2\cos50^{°}\sin25^{°} +2\cos60^{°} \sin25^{°}] \\
& = \frac{1}{4}[ 2\cos50^{°}\sin25^{°} +2 \frac{1}{2} \sin25^{°}] \\
& = \frac{1}{4}[ 2\cos50^{°}\sin25^{°} +\sin25^{°}] \\
& = \frac{1}{4} [ \sin75^{°} -\sin 25^{°} +\sin25^{°}] \\
& =\frac{1}{4}  \sin75^{°}
\end{align}
</s>