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.
Best Answer
Note that $\sin(95^\circ)=\sin(85^\circ)$.
We can get an explicit formula for $\sin(75^\circ)$ in various ways, probably most simply by writing it as $\sin(30^\circ+45^\circ)$ and using the Addition Law.