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If $x=\cos 10^{\circ} \cos 20^{\circ} \cos 40^{\circ}$, then the value of $x$ is
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Verified Answer
The correct answer is:
$\frac{1}{8} \cot 10^{\circ}$
$\begin{aligned}
& x=\cos 10^{\circ} \cos 20^{\circ} \cos 40^{\circ} \\
&=\frac{1}{2 \sin 10^{\circ}}\left[2 \sin 10^{\circ} \cos 10^{\circ} \cos 20^{\circ} \cos 40^{\circ}\right] \\
&=\frac{1}{2.2 \sin 10^{\circ}}\left[2 \sin 20^{\circ} \cos 20^{\circ} \cos 40^{\circ}\right] \\
&=\frac{1}{2.4 \sin 10^{\circ}}\left[2 \sin 40^{\circ} \cos 40^{\circ}\right)=\frac{1}{8 \sin 10^{\circ}}\left(\sin 80^{\circ}\right) \\
&=\frac{1}{8 \sin 10^{\circ}} \cos 10^{\circ}=\frac{1}{8} \cot 10^{\circ}
\end{aligned}$
& x=\cos 10^{\circ} \cos 20^{\circ} \cos 40^{\circ} \\
&=\frac{1}{2 \sin 10^{\circ}}\left[2 \sin 10^{\circ} \cos 10^{\circ} \cos 20^{\circ} \cos 40^{\circ}\right] \\
&=\frac{1}{2.2 \sin 10^{\circ}}\left[2 \sin 20^{\circ} \cos 20^{\circ} \cos 40^{\circ}\right] \\
&=\frac{1}{2.4 \sin 10^{\circ}}\left[2 \sin 40^{\circ} \cos 40^{\circ}\right)=\frac{1}{8 \sin 10^{\circ}}\left(\sin 80^{\circ}\right) \\
&=\frac{1}{8 \sin 10^{\circ}} \cos 10^{\circ}=\frac{1}{8} \cot 10^{\circ}
\end{aligned}$
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