Trigonometry

Geometry Level 3

If A = 2 π 7 A=\frac {2\pi}7 , what is cos ⁡ A cos ⁡ 2 A cos ⁡ 4 A = ? \cos A \cos 2A \cos 4A = ?

1 2 \frac 12 1 4 \frac 14 1 1 1 8 \frac 18

Dipanjoy Saha by dipanjoy saha , 20, India

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1 solution

Chew-Seong Cheong
Aug 19, 2017

X = cos ⁡ 2 π 7 cos ⁡ 4 π 7 cos ⁡ 8 π 7 = sin ⁡ 2 π 7 cos ⁡ 2 π 7 cos ⁡ 4 π 7 cos ⁡ 8 π 7 sin ⁡ 2 π 7 = sin ⁡ 4 π 7 cos ⁡ 4 π 7 cos ⁡ 8 π 7 2 sin ⁡ 2 π 7 = sin ⁡ 8 π 7 cos ⁡ 8 π 7 4 sin ⁡ 2 π 7 = sin ⁡ 16 π 7 8 sin ⁡ 2 π 7 Note that sin ⁡ ( π − x ) = sin ⁡ x = sin ⁡ 2 π 7 8 sin ⁡ 2 π 7 = 1 8 \begin{aligned} X & = \cos \frac {2\pi}7 \cos \frac {4\pi}7 \cos \frac {8\pi}7 \\ & = \frac {{\color{#3D99F6}\sin \frac {2\pi}7}\cos \frac {2\pi}7 \cos \frac {4\pi}7 \cos \frac {8\pi}7}{\color{#3D99F6}\sin \frac {2\pi}7} \\ & = \frac {{\color{#3D99F6}\sin \frac {4\pi}7}\cos \frac {4\pi}7 \cos \frac {8\pi}7}{{\color{#3D99F6}2}\sin \frac {2\pi}7} \\ & = \frac {{\color{#3D99F6}\sin \frac {8\pi}7}\cos \frac {8\pi}7}{{\color{#3D99F6}4}\sin \frac {2\pi}7} \\ & = \frac {{\color{#3D99F6}\sin \frac {16\pi}7}}{{\color{#3D99F6}8}\sin \frac {2\pi}7} & \small \color{#3D99F6} \text{Note that }\sin (\pi -x) = \sin x \\ & = \frac {{\color{#3D99F6}\sin \frac {2\pi}7}}{8\sin \frac {2\pi}7} \\ & = \boxed{\dfrac 18} \end{aligned}

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