Find the value

Geometry Level 3

cos ⁡ 2 0 ∘ cos ⁡ 4 0 ∘ cos ⁡ 6 0 ∘ cos ⁡ 8 0 ∘ = ? \large \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos80^\circ = \, ?


The answer is 0.0625.

Atul Kumar Ashish by Atul Kumar Ashish , 21, India

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2 solutions

Let x = cos ⁡ ( 2 0 ∘ ) cos ⁡ ( 4 0 ∘ ) cos ⁡ ( 8 0 ∘ ) x = \cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ}) . Then using the identity 1 2 sin ⁡ ( 2 x ) = sin ⁡ ( x ) cos ⁡ ( x ) \dfrac{1}{2}\sin(2x) = \sin(x)\cos(x) we see that

x sin ⁡ ( 2 0 ∘ ) = 1 2 sin ⁡ ( 4 0 ∘ ) cos ⁡ ( 4 0 ∘ ) cos ⁡ ( 8 0 ∘ ) = 1 4 sin ⁡ ( 8 0 ∘ ) cos ⁡ ( 8 0 ∘ ) = 1 8 sin ⁡ ( 16 0 ∘ ) = 1 8 sin ⁡ ( 2 0 ∘ ) x\sin(20^{\circ}) = \dfrac{1}{2}\sin(40^{\circ})\cos(40^{\circ})\cos(80^{\circ}) = \dfrac{1}{4}\sin(80^{\circ})\cos(80^{\circ}) = \dfrac{1}{8}\sin(160^{\circ}) = \dfrac{1}{8}\sin(20^{\circ})

since sin ⁡ ( y ) = sin ⁡ ( 18 0 ∘ − y ) \sin(y) =\sin(180^{\circ} - y) . Thus x = 1 8 x = \dfrac{1}{8} , and so x cos ⁡ ( 6 0 ∘ ) = 1 8 ∗ 1 2 = 1 16 = 0.0625 x\cos(60^{\circ}) = \dfrac{1}{8}*\dfrac{1}{2} = \dfrac{1}{16} = \boxed{0.0625} .

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Nice way, I didn't notice the angles doubling so I used,
cos ⁡ ( θ ) cos ⁡ ( 6 0 ∘ − θ ) cos ⁡ ( 6 0 ∘ + θ ) = cos ⁡ ( 3 θ ) 4 \cos(\theta) \cos(60^\circ - \theta) \cos(60^\circ + \theta) = \dfrac{\cos(3\theta)}{4}

A Former Brilliant Member - 5 years, 1 month ago
Otto Bretscher
Apr 22, 2016

This is Morrie's Law (so called by Richard Feynman) divided by 2. The answer is 1 16 = 0.0625 \frac{1}{16}=\boxed{0.0625}

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