Trigonometry (4)

Geometry Level 4

cos ⁡ π 65 cos ⁡ 2 π 65 cos ⁡ 4 π 65 cos ⁡ 8 π 65 cos ⁡ 16 π 65 cos ⁡ 32 π 65 \cos \dfrac{π}{65} \cos \dfrac{2π}{65} \cos \dfrac{4π}{65} \cos \dfrac{8π}{65} \cos \dfrac{16π}{65} \cos \dfrac{32π}{65}

Find the value of the expression above to 4 decimal places.


The answer is 0.0156.

Rahil Sehgal by Rahil Sehgal , 20, India

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1 solution

Chew-Seong Cheong
Apr 18, 2017

Consider the following:

P = cos ⁡ π 65 cos ⁡ 2 π 65 cos ⁡ 4 π 65 cos ⁡ 8 π 65 cos ⁡ 16 π 65 cos ⁡ 32 π 65 = sin ⁡ π 65 cos ⁡ π 65 cos ⁡ 2 π 65 cos ⁡ 4 π 65 cos ⁡ 8 π 65 cos ⁡ 16 π 65 cos ⁡ 32 π 65 sin ⁡ π 65 = sin ⁡ 2 π 65 cos ⁡ 2 π 65 cos ⁡ 4 π 65 cos ⁡ 8 π 65 cos ⁡ 16 π 65 cos ⁡ 32 π 65 2 sin ⁡ π 65 = sin ⁡ 4 π 65 cos ⁡ 4 π 65 cos ⁡ 8 π 65 cos ⁡ 16 π 65 cos ⁡ 32 π 65 4 sin ⁡ π 65 = sin ⁡ 8 π 65 cos ⁡ 8 π 65 cos ⁡ 16 π 65 cos ⁡ 32 π 65 8 sin ⁡ π 65 = sin ⁡ 16 π 65 cos ⁡ 16 π 65 cos ⁡ 32 π 65 16 sin ⁡ π 65 = sin ⁡ 32 π 65 cos ⁡ 32 π 65 32 sin ⁡ π 65 = sin ⁡ 64 π 65 64 sin ⁡ π 65 = sin ⁡ ( π − π 65 ) 64 sin ⁡ π 65 Note that sin ⁡ ( π − x ) = sin ⁡ x = sin ⁡ π 65 64 sin ⁡ π 65 = 1 64 = 0.015625 \begin{aligned} P & = \cos \frac \pi {65} \cos \frac {2\pi}{65} \cos \frac {4\pi}{65} \cos \frac {8\pi}{65} \cos \frac {16\pi}{65} \cos \frac {32\pi}{65} \\ & = \frac {{\color{#3D99F6}\sin \frac \pi {65}} \cos \frac \pi {65} \cos \frac {2\pi}{65} \cos \frac {4\pi}{65} \cos \frac {8\pi}{65} \cos \frac {16\pi}{65} \cos \frac {32\pi}{65}}{\color{#3D99F6}\sin \frac \pi {65}} \\ & = \frac {{\color{#3D99F6}\sin \frac {2\pi}{65}} \cos \frac {2\pi}{65} \cos \frac {4\pi}{65} \cos \frac {8\pi}{65} \cos \frac {16\pi}{65} \cos \frac {32\pi}{65}}{{\color{#3D99F6}2} \sin \frac \pi {65}} \\ & = \frac {{\color{#3D99F6}\sin \frac {4\pi}{65}} \cos \frac {4\pi}{65} \cos \frac {8\pi}{65} \cos \frac {16\pi}{65} \cos \frac {32\pi}{65}}{{\color{#3D99F6}4} \sin \frac \pi {65}} \\ & = \frac {{\color{#3D99F6}\sin \frac {8\pi}{65}} \cos \frac {8\pi}{65} \cos \frac {16\pi}{65} \cos \frac {32\pi}{65}}{{\color{#3D99F6}8} \sin \frac \pi {65}} \\ & = \frac {{\color{#3D99F6}\sin \frac {16\pi}{65}} \cos \frac {16\pi}{65} \cos \frac {32\pi}{65}}{{\color{#3D99F6}16} \sin \frac \pi {65}} \\ & = \frac {{\color{#3D99F6}\sin \frac {32\pi}{65}} \cos \frac {32\pi}{65}}{{\color{#3D99F6}32} \sin \frac \pi {65}} \\ & = \frac {\color{#3D99F6}\sin \frac {64\pi}{65}}{{\color{#3D99F6}64} \sin \frac \pi {65}} \\ & = \frac {\color{#3D99F6}\sin \left(\pi - \frac \pi{65}\right)}{64 \sin \frac \pi {65}} \quad \quad \small \color{#3D99F6} \text{Note that }\sin (\pi-x) = \sin x \\ & = \frac {\color{#3D99F6}\sin \frac \pi{65}}{64 \sin \frac \pi {65}} = \frac 1{64} = \boxed{0.015625} \end{aligned}

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