# Proving the trigonometric identity

Please help me in proving the following idenity:

$$8\cdot \cos 40^\circ\cdot \cos 20^\circ \cdot \cos 10^\circ = \cot 10^\circ$$

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Do you read the comments on your other questions? If you do, you know what the attitude of doing nothing about them looks like. – Did Sep 16 '12 at 18:38
Thank you for taking the time to do what the community expects you to do, Ha Hi. – Gigili Sep 16 '12 at 19:31

## 2 Answers

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}

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Hint: Multiply the expression on the left-hand side by $\sin(10^\circ)$, and use repeatedly the identity $\sin 2x=2\sin x\cos x$.

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