Trigonometry problem no. Ek

Geometry Level 4

1 4 cos ⁡ 1 2 ∘ cos ⁡ 2 4 ∘ cos ⁡ 3 6 ∘ cos ⁡ 4 8 ∘ cos ⁡ 7 2 ∘ cos ⁡ 8 4 ∘ = ? \frac{1}{4\cos{12^\circ}\cos{24^\circ}\cos{36^\circ}\cos{48^\circ}\cos{72^\circ}\cos{84^\circ}} = \, ?


The answer is 16.

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Anurag Pandey by Anurag Pandey , 23, India

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1 solution

Chew-Seong Cheong
Jan 10, 2017

X = 1 4 cos ⁡ 1 2 ∘ cos ⁡ 2 4 ∘ cos ⁡ 3 6 ∘ cos ⁡ 4 8 ∘ cos ⁡ 7 2 ∘ cos ⁡ 8 4 ∘ = sin ⁡ 1 2 ∘ sin ⁡ 2 4 ∘ sin ⁡ 3 6 ∘ sin ⁡ 4 8 ∘ sin ⁡ 7 2 ∘ sin ⁡ 8 4 ∘ 4 sin ⁡ 1 2 ∘ cos ⁡ 1 2 ∘ sin ⁡ 2 4 ∘ cos ⁡ 2 4 ∘ sin ⁡ 3 6 ∘ cos ⁡ 3 6 ∘ sin ⁡ 4 8 ∘ cos ⁡ 4 8 ∘ sin ⁡ 7 2 ∘ cos ⁡ 7 2 ∘ sin ⁡ 8 4 ∘ cos ⁡ 8 4 ∘ = 2 6 sin ⁡ 1 2 ∘ sin ⁡ 2 4 ∘ sin ⁡ 3 6 ∘ sin ⁡ 4 8 ∘ sin ⁡ 7 2 ∘ sin ⁡ 8 4 ∘ 4 sin ⁡ 2 4 ∘ sin ⁡ 4 8 ∘ sin ⁡ 7 2 ∘ sin ⁡ 9 6 ∘ sin ⁡ 14 4 ∘ sin ⁡ 16 8 ∘ = \require c a n c e l 16 sin ⁡ 1 2 ∘ sin ⁡ 2 4 ∘ sin ⁡ 3 6 ∘ sin ⁡ 4 8 ∘ sin ⁡ 7 2 ∘ sin ⁡ 8 4 ∘ sin ⁡ 2 4 ∘ sin ⁡ 4 8 ∘ sin ⁡ 7 2 ∘ sin ⁡ 9 6 ∘ sin ⁡ 14 4 ∘ sin ⁡ 16 8 ∘ = 16 sin ⁡ 1 2 ∘ sin ⁡ 3 6 ∘ sin ⁡ 8 4 ∘ sin ⁡ 9 6 ∘ sin ⁡ 14 4 ∘ sin ⁡ 16 8 ∘ = 16 sin ⁡ 1 2 ∘ sin ⁡ 3 6 ∘ sin ⁡ 8 4 ∘ sin ⁡ ( 180 − 9 6 ∘ ) sin ⁡ ( 18 0 ∘ − 14 4 ∘ ) sin ⁡ ( 18 0 ∘ − 16 8 ∘ ) = 16 sin ⁡ 1 2 ∘ sin ⁡ 3 6 ∘ sin ⁡ 8 4 ∘ sin ⁡ 8 4 ∘ sin ⁡ 3 6 ∘ sin ⁡ 1 2 ∘ = \require c a n c e l 16 sin ⁡ 1 2 ∘ sin ⁡ 3 6 ∘ sin ⁡ 8 4 ∘ sin ⁡ 8 4 ∘ sin ⁡ 3 6 ∘ sin ⁡ 1 2 ∘ = 16 \begin{aligned} X & = \frac 1{4 \cos 12^\circ \cos 24^\circ \cos 36^\circ \cos 48^\circ \cos 72^\circ \cos 84^\circ} \\ & = \frac {\color{#3D99F6} \sin 12^\circ \sin 24^\circ \sin 36^\circ \sin 48^\circ \sin 72^\circ \sin 84^\circ}{4 {\color{#3D99F6} \sin 12^\circ} \cos 12^\circ {\color{#3D99F6} \sin 24^\circ} \cos 24^\circ {\color{#3D99F6} \sin 36^\circ} \cos 36^\circ {\color{#3D99F6} \sin 48^\circ} \cos 48^\circ {\color{#3D99F6} \sin 72^\circ} \cos 72^\circ {\color{#3D99F6} \sin 84^\circ} \cos 84^\circ} \\ & = \frac {2^6 \sin 12^\circ \sin 24^\circ \sin 36^\circ \sin 48^\circ \sin 72^\circ \sin 84^\circ}{4 \sin 24^\circ \sin 48^\circ \sin 72^\circ \sin 96^\circ \sin 144^\circ \sin 168^\circ} \\ & = \require {cancel} \frac {16 \sin 12^\circ \cancel {\sin 24^\circ} \sin 36^\circ \cancel {\sin 48^\circ} \cancel {\sin 72^\circ} \sin 84^\circ}{\cancel {\sin 24^\circ} \cancel {\sin 48^\circ} \cancel {\sin 72^\circ} \sin 96^\circ \sin 144^\circ \sin 168^\circ} \\ & = \frac {16 \sin 12^\circ \sin 36^\circ \sin 84^\circ}{\sin 96^\circ \sin 144^\circ \sin 168^\circ} \\ & = \frac {16 \sin 12^\circ \sin 36^\circ \sin 84^\circ}{\sin (180-96^\circ) \sin (180^\circ - 144^\circ) \sin (180^\circ -168^\circ)} \\ & = \frac {16 \sin 12^\circ \sin 36^\circ \sin 84^\circ}{\sin 84^\circ \sin 36^\circ \sin 12^\circ} \\ & = \require {cancel} \frac {16 \cancel {\sin 12^\circ} \cancel {\sin 36^\circ} \cancel {\sin 84^\circ}}{\cancel {\sin 84^\circ} \cancel {\sin 36^\circ} \cancel {\sin 12^\circ}} \\ & = \boxed{16} \end{aligned}

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