Prove that cos 6⁰ cos 42⁰ cos 66⁰ cos 78⁰=1/16 .

Question

Prove that [latex] \cos 6^{\circ} \cos 42^{\circ} \cos 66^{\circ} \cos 78^{\circ}=\frac{1}{16} [/latex].

Accepted Answer

We have

[latex]\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}\ \[/latex]

[latex]\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}[/latex]

Question 15.4.24: Prove that cos 6⁰ cos 42⁰ cos 66⁰ cos 78⁰=1/16 ....

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