Proving the trigonometric identity Please help me in proving the following idenity:
$$8\cdot \cos 40^\circ\cdot \cos 20^\circ \cdot \cos 10^\circ = \cot 10^\circ$$
 A: Hint: Multiply the expression on the left-hand side by $\sin(10^\circ)$, and use repeatedly the identity $\sin 2x=2\sin x\cos x$.
A: Before proving the trigonometrics relation note some trigonometrics formula which will utilize during the proof.
a) $2\cos\alpha\cdot\sin\alpha=\sin 2\alpha$ 
b) $\sin\alpha=\cos(90^\circ-\alpha)$
c) $\frac{\cos\alpha}{\sin\alpha}=\cot\alpha$
\begin{align}
8\cdot \cos &40^0\cdot \cos 20^0 \cdot \cos 10^0\\
& = \frac{8\cdot \cos 40^0\cdot \cos 20^0 \cdot \cos 10^0\cdot \sin 10^0}{\sin 10^0}\\
& =\frac{4\cdot \cos 40^0\cdot \cos 20^0 \cdot 2\cos 10^0\cdot \sin 10^0}{\sin 10^0}\\
& =\frac{4\cdot \cos 40^0\cdot \cos 20^0 \cdot \sin 2\cdot 10^0}{\sin 10^0}\\
&=\frac{2\cdot \cos 40^0\cdot 2\cos 20^0 \cdot \sin 20^0}{\sin 10^0}\\
&=\frac{2\cdot \cos 40^0\cdot \sin 2\cdot 20^0}{\sin 10^0}\\
&=\frac{2\cdot \cos 40^0\cdot\sin 40^0}{\sin 10^0}\\
&=\frac{\sin 2\cdot 40^0}{\sin 10^0}\\
& =\frac{\sin 80^0}{\sin 10^0}\\
& = \frac{\cos 10^0}{\sin 10^0} = \cot 10^0
\end{align}
