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Prove that: \(\cos 20^{\circ }\cos 40^{\circ }\cos 60^{\circ }\cos 80^{o}=\frac {1}{16}\)
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Prove that: \(\cos 20^{\circ }\cos 40^{\circ }\cos 60^{\circ }\cos 80^{o}=\frac {1}{16}\)

Answer

\(\text{LHS}=\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}\)
\(\Rightarrow \quad \text{LHS} =\cos 60^{\circ}\left(\cos 20^{\circ} \cos 40^{\circ}\right) \cos 80^{\circ}\)
\(\Rightarrow \quad \text{LHS}=\frac{1}{2} \times \frac{1}{2}\left(2 \cos 20^{\circ} \cos 40^{\circ}\right) \cos 80^{\circ}\quad\left[\because \cos 60 ^\circ=\frac{1}{2}\right]\)
\(\Rightarrow\quad \text{LHS} =\frac{1}{4}\left[\left\{\cos \left(40^{\circ}+20^{\circ}\right)+\cos \left(40^{\circ}-20^{\circ}\right)\right\} \cos 80^{\circ}\right] \quad\left[\begin{array}{l}\because 2 \cos A \cos B \\ =\cos (A+B)+\cos (A-B)]\end{array}\right.\)
\(\Rightarrow \quad \text{LHS}=\frac{1}{4}\left\{\left(\cos 60^{\circ}+\cos 20^{\circ}\right) \cos 80^{\circ}\right\}\)
\(\Rightarrow \quad \mathrm{LHS}=\frac{1}{4}\left\{\left(\frac{1}{2}+\cos 20^{\circ}\right) \cos 80^{\circ}\right\}\)
\(\Rightarrow\quad \text{LHS}=\frac{1}{4}\left\{\frac{1}{2} \cos 80^{\circ}+\cos 80^{\circ} \cos 20^{\circ}\right\}\)
\(\Rightarrow\quad \text{LHS}=\frac{1}{8}\left\{\cos 80^{\circ}+2\cos 80^{\circ} \cos 20^{\circ}\right\}\)
\(\Rightarrow \quad \text{LHS}=\frac{1}{8}\left[\cos 80^{\circ}+\left\{\cos \left(80^{\circ}+20^{\circ}\right)+\cos \left(80^{\circ}-20^{\circ}\right)\right\}\right]\left[\begin{array}{l}\because 2 \cos A \cos B \\ =\cos (A+B)+\cos (A-B)\end{array}\right]\)
\(\Rightarrow \quad \text{LHS}=\frac{1}{8}\left\{\cos 80^{\circ}+\cos 100^{\circ}+\cos 60^{\circ}\right\}\)
\(\Rightarrow\quad \text{LHS}=\frac{1}{8}\left\{\cos 80^{\circ}+\cos \left(180^{\circ}-80^{\circ}\right)+\cos 60^{\circ}\right\}\)
\(\Rightarrow \quad \mathrm{LHS}=\frac{1}{8}\left\{\cos 80^{\circ}-\cos 80^{\circ}+\cos 60^{\circ}\right\} \quad \left[\begin{array}{c}\therefore \cos \left(180^{\circ}-x\right)=-\cos x \quad \therefore \cos \left(180^{\circ}-80^{\circ}\right)=-\cos 80^{\circ}\end{array}\right]\)
\(\Rightarrow \quad \text{LHS} =\frac{1}{8}\left\{\cos 80^{\circ}-\cos 80^{\circ}+\frac{1}{2}\right\}=\frac{1}{8} \times \frac{1}{2}=\frac{1}{16}=\mathrm{RHS}\)
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