A geometry problem by Jose Sacramento

Geometry Level 4

cos ⁡ ( 2 π 7 ) + cos ⁡ ( 4 π 7 ) + cos ⁡ ( 6 π 7 ) cos ⁡ ( π 7 ) × cos ⁡ ( 3 π 7 ) × cos ⁡ ( 5 π 7 ) = ? \large \frac{\cos\left( \frac{2\pi}7 \right) + \cos\left( \frac{4\pi}7 \right) + \cos\left( \frac{6\pi}7 \right) }{\cos\left( \frac{\pi}7 \right) \times \cos\left( \frac{3\pi}7 \right) \times \cos\left( \frac{5\pi}7 \right)} = \, ?


The answer is 4.

Jose Sacramento by Jose Sacramento , 66, Portugal

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1 solution

cos ⁡ ( π − A ) = − cos ⁡ A ∴ cos ⁡ ( 2 π 7 ) + cos ⁡ ( 4 π 7 ) + cos ⁡ ( 6 π 7 ) = cos ⁡ ( 2 π 7 ) − cos ⁡ ( 3 π 7 ) − cos ⁡ ( π 7 ) = − { cos ⁡ ( π 7 ) − cos ⁡ ( 2 π 7 ) + cos ⁡ ( 3 π 7 ) } = − 1 2 \cos( π-A) = - \cos A \\ \therefore\ \cos\left( \dfrac{2\pi}7 \right) + \cos\left( \dfrac{4\pi}7 \right) + \cos\left( \dfrac{6\pi}7 \right)\\ =\cos\left( \dfrac{2\pi}7 \right) - \cos\left( \dfrac{3\pi}7 \right) - \cos\left( \dfrac{\pi}7 \right)\\ =- \left\{ \color{#3D99F6}{\cos\left( \dfrac{\pi}7 \right) - \cos\left( \dfrac{2\pi}7 \right) + \cos\left( \dfrac{3\pi}7 \right) }\right \}\\ =- \color{#3D99F6}{\dfrac 1 2}\\ \ \ \ \\
A g a i n cos ⁡ ( π 7 ) × cos ⁡ ( 3 π 7 ) × cos ⁡ ( 5 π 7 ) Again\ \ \cos\left( \dfrac{\pi}7 \right) \times \cos\left( \dfrac{3\pi}7 \right) \times \cos\left( \dfrac{5\pi}7 \right)\\
= cos ⁡ ( π 7 ) × cos ⁡ ( 3 π 7 ) × ( − cos ⁡ ( 2 π 7 ) ) =\cos\left( \dfrac{\pi}7 \right) \times \cos\left( \dfrac{3\pi}7 \right) \times \left (- \cos\left( \dfrac{2\pi}7 \right) \right)\\
= − { cos ⁡ ( π 7 ) × cos ⁡ ( 2 π 7 ) × cos ⁡ ( 3 π 7 ) } =- \left \{\color{#3D99F6}{ \cos\left( \dfrac{\pi}7 \right) \times \cos\left( \dfrac{2\pi}7 \right) \times \cos\left( \dfrac{3\pi}7 \right) } \right \}\\


= − 1 8 =-\color{#3D99F6}{\dfrac 1 8}\\
\thereforr / f = cos ⁡ ( 2 π 7 ) + cos ⁡ ( 4 π 7 ) + cos ⁡ ( 6 π 7 ) cos ⁡ ( π 7 ) × cos ⁡ ( 3 π 7 ) × cos ⁡ ( 5 π 7 ) = 4. \thereforr/ \Large f=\dfrac{\cos\left( \dfrac{2\pi}7 \right) + \cos\left( \dfrac{4\pi}7 \right) + \cos\left( \dfrac{6\pi}7 \right) }{\cos\left( \dfrac{\pi}7 \right) \times \cos\left( \dfrac{3\pi}7 \right) \times \cos\left( \dfrac{5\pi}7 \right)} = \ \ \ \color{#D61F06}{4}. \\ \ \ \\
\ \ \ \\ \\ \ \ \\

Proofs of blue formulae can be had from Google or yahoo. Below is one proof.

cos ⁡ ( π 7 ) − cos ⁡ ( 2 π 7 ) + cos ⁡ ( 3 π 7 ) = 1 2 . \cos\left( \dfrac{\pi}7 \right) - \cos\left( \dfrac{2\pi}7 \right) + \cos\left( \dfrac{3\pi}7 \right)=\dfrac 1 2. \\
Multuiply and divide numerator of f by 2sin(2π}7 ,.......
and since sin(2A) = 2sinAcosA ... and 2sinAcosB = sin(A+B) + sin(A-B).....sin( π - A)=sinA.
Numerator of f
= 1/2sin(π/7) { 2sin(π/7)cos(π/7) + 2sin(π/7)cos(3π/7) + 2sin(π/7)cos(5π/7) }
= 1/2sin(π/7){ sin(2π/7) + sin(π/7+3π/7) + sin(π/7-3π/7) + sin(π/7+5π/7) + sin(π/7-5π/7) }
= 1/2sin(π/7){ sin(2π/7) + sin(4π/7) + sin(-2π/7) + sin(6π/7) + sin(-4π/7) }
= {sin(π/7)}/2sin(π/7)
= 1/2 .

cos ⁡ ( π 7 ) × cos ⁡ ( 2 π 7 ) × cos ⁡ ( 3 π 7 ) = 1 8 . \cos\left( \dfrac{\pi}7 \right) \times \cos\left( \dfrac{2\pi}7 \right) \times \cos\left( \dfrac{3\pi}7 \right)=\dfrac 1 8.

Muliply denominator of f by 8 sin pi/7 ....... and sin( π - A)=sinA ...we get,
8 sin pi/7 cos pi/7 cos 2pi/7 cos 3pi/7 = 4(2 sin pi/7 cos pi/7 cos 2pi/7 cos 3pi/7) = 4( sin 2pi/7 cos 2pi/7 cos 3pi/7) ..........as sin 2A = 2 sin A cos A.
= 2( 2 sin 2pi/7 cos 2pi/7 cos 3pi/7)
= 2 ( sin 4pi/7 cos 3pi/7) = - 2 ( sin 4pi/7 cos 4pi/7 )...................as cos 3pi/7 = - cos 4pi/7 = - sin 8pi/7 = sin pi/7 .
So denominator of f
=cos pi/7 cos 2pi/7 cos 3pi/7 = 1/8.

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