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Question 15.4.24: Prove that cos 6⁰ cos 42⁰ cos 66⁰ cos 78⁰=1/16 ....
Senior Secondary School Mathematics for Class 11 [1185468]
Prove that
\cos 6^{\circ} \cos 42^{\circ} \cos 66^{\circ} \cos 78^{\circ}=\frac{1}{16}
.
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We have
\begin{aligned}\text { LHS } & =\cos 6^{\circ} \cos 42^{\circ} \cos 66^{\circ} \cos 78^{\circ} \\ \\& =\frac{1}{4}\left(2 \cos 66^{\circ} \cos 6^{\circ}\right)\left(2 \cos 78^{\circ} \cos 42^{\circ}\right)\\ \\& =\frac{1}{4}\left[\cos \left(66^{\circ}+6^{\circ}\right)+\cos \left(66^{\circ}-6^{\circ}\right)\right] \times \\ \\ & [\cos (78^{\circ}+42^{\circ})+\cos (78^{\circ}-42^{\circ})] \end{aligned}\\ \\
\begin{aligned} &=\frac{1}{4}(\cos 72^{\circ}+ \cos 60^{\circ})(\cos 120^{\circ}+ \cos 36^{\circ}) \\ \\& =\frac{1}{4}\left(\sin 18^{\circ}+\frac{1}{2}\right)\left(-\frac{1}{2}+\cos 36^{\circ}\right)\\ \\ & [ \because \cos 72^{\circ} = \cos ( 90^{\circ} -18^{\circ}) = \sin 18^{\circ}] \\ \\& =\frac{1}{4}\left[\frac{(\sqrt{5}-1)}{4}+\frac{1}{2}\right]\left[-\frac{1}{2}+\frac{(\sqrt{5}+1)}{4}\right]\\ \\ & \left[ \because \sin 18^{\circ} =\frac{(\sqrt{5}-1)}{4} \text{ and } \cos 36^{\circ} =\frac{(\sqrt{5}+1)}{4} \right] \\ \\& =\frac{1}{4} \cdot \frac{(\sqrt{5}+1)}{4} \cdot \frac{(\sqrt{5}-1)}{4}=\frac{(5-1)}{64} \\ \\& =\frac{4}{64}=\frac{1}{16}=\text { RHS. }\end{aligned}
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