Trigonometry (1)

Geometry Level 3

cos ⁡ 2 π 15 cos ⁡ 4 π 15 cos ⁡ 8 π 15 cos ⁡ 16 π 15 \large \cos \dfrac{2π}{15} \ \cos \dfrac{4π}{15} \ \cos \dfrac{8π}{15} \ \cos \dfrac{16π}{15}

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


The answer is 0.0625.

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Rahil Sehgal by Rahil Sehgal , 20, India

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2 solutions

Md Zuhair
Apr 15, 2017

cos ⁡ 2 π 15 ⋅ cos ⁡ 4 π 15 ⋅ cos ⁡ 8 π 15 ⋅ cos ⁡ 16 π 15 \cos \dfrac{2π}{15} \cdot \cos \dfrac{4π}{15} \cdot \cos \dfrac{8π}{15} \cdot \cos \dfrac{16π}{15} can be reduced by using multiple angle formula sin ⁡ 2 θ = 2 sin ⁡ θ cos ⁡ θ \sin 2 \theta = 2\sin \theta \cos \theta .

Ok so our expression is cos ⁡ 2 π 15 ⋅ cos ⁡ 4 π 15 ⋅ cos ⁡ 8 π 15 ⋅ cos ⁡ 16 π 15 \cos \dfrac{2π}{15} \cdot \cos \dfrac{4π}{15} \cdot \cos \dfrac{8π}{15} \cdot \cos \dfrac{16π}{15} Take θ = π 15 \theta = \dfrac{\pi}{15}

Now

cos ⁡ 2 π 15 ⋅ cos ⁡ 4 π 15 ⋅ cos ⁡ 8 π 15 ⋅ cos ⁡ 16 π 15 \cos \dfrac{2π}{15} \cdot \cos \dfrac{4π}{15} \cdot \cos \dfrac{8π}{15} \cdot \cos \dfrac{16π}{15}

⟹ cos ⁡ 2 θ ⋅ cos ⁡ 4 θ ⋅ cos ⁡ 8 θ ⋅ cos ⁡ 16 θ \implies \cos {2 \theta} \cdot \cos {4 \theta} \cdot \cos {8\theta} \cdot \cos {16 \theta}

⟹ 1 2 sin ⁡ 2 θ 2 sin ⁡ 2 θ × cos ⁡ 2 θ ⋅ cos ⁡ 4 θ ⋅ cos ⁡ 8 θ ⋅ cos ⁡ 16 θ \implies \dfrac{1}{2 \sin 2\theta} 2\sin 2\theta \times \cos {2 \theta} \cdot \cos {4 \theta} \cdot \cos {8\theta} \cdot \cos {16 \theta}

⟹ 1 2 sin ⁡ 2 θ sin ⁡ 4 θ ⋅ cos ⁡ 4 θ ⋅ cos ⁡ 8 θ ⋅ cos ⁡ 16 θ \implies \dfrac{1}{2 \sin 2\theta} \sin 4\theta \cdot \cos {4 \theta} \cdot \cos {8\theta} \cdot \cos {16 \theta}

⟹ 1 2 2 sin ⁡ 2 θ 2 sin ⁡ 4 θ ⋅ cos ⁡ 4 θ ⋅ cos ⁡ 8 θ ⋅ cos ⁡ 16 θ \implies \dfrac{1}{2^2 \sin 2\theta } 2\sin 4\theta \cdot \cos {4 \theta} \cdot \cos {8\theta} \cdot \cos {16 \theta}

⟹ 1 2 2 sin ⁡ 2 θ sin ⁡ 8 θ ⋅ cos ⁡ 8 θ ⋅ cos ⁡ 16 θ \implies \dfrac{1}{2^2 \sin 2\theta } \sin 8\theta \cdot \cos {8\theta} \cdot \cos {16 \theta}

⟹ 1 2 3 sin ⁡ 2 θ 2 sin ⁡ 8 θ ⋅ cos ⁡ 8 θ ⋅ cos ⁡ 16 θ \implies \dfrac{1}{2^3 \sin 2\theta } 2\sin 8\theta \cdot \cos {8\theta} \cdot \cos {16 \theta}

⟹ 1 2 3 sin ⁡ 2 θ sin ⁡ 16 θ ⋅ cos ⁡ 16 θ \implies \dfrac{1}{2^3 \sin 2\theta } \sin 16\theta \cdot \cos {16 \theta}

⟹ 1 2 4 sin ⁡ 2 θ 2 sin ⁡ 16 θ ⋅ cos ⁡ 16 θ \implies \dfrac{1}{2^4 \sin 2\theta } 2\sin 16\theta \cdot \cos {16 \theta}

⟹ 1 2 4 sin ⁡ 2 θ sin ⁡ 32 θ \implies \dfrac{1}{2^4 \sin 2\theta } \sin 32\theta

Now putting value of θ = π 15 \theta = \dfrac{\pi}{15} , we have,

⟹ 1 2 4 sin ⁡ 2 π 15 sin ⁡ 32 π 15 = P ( S a y ) \implies \dfrac{1}{2^4 \sin \dfrac{2\pi}{15}} \sin \dfrac{32\pi}{15} = P(Say)

Now sin ⁡ 32 π 15 ⟹ sin ⁡ ( 2 π + π 15 ) ⟹ sin ⁡ π 15 \sin \dfrac{32\pi}{15} \implies \sin (2\pi+ \dfrac{\pi}{15}) \implies \sin \dfrac{\pi}{15}

P reduces to

1 2 4 ⟹ 1 16 ⟹ 0.0625 ( A n s ) \dfrac{1}{2^4} \implies \dfrac{1}{16} \implies \boxed{0.0625}(Ans)

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Thanks you very much... (+1)

Rahil Sehgal - 4 years, 1 month ago

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Welcome very much

Md Zuhair - 4 years, 1 month ago
Chew-Seong Cheong
Apr 17, 2017

P = cos ⁡ 2 π 15 cos ⁡ 4 π 15 cos ⁡ 8 π 15 cos ⁡ 16 π 15 = sin ⁡ 2 π 15 cos ⁡ 2 π 15 cos ⁡ 4 π 15 cos ⁡ 8 π 15 cos ⁡ 16 π 15 sin ⁡ 2 π 15 = sin ⁡ 4 π 15 cos ⁡ 4 π 15 cos ⁡ 8 π 15 cos ⁡ 16 π 15 2 sin ⁡ 2 π 15 = sin ⁡ 8 π 15 cos ⁡ 8 π 15 cos ⁡ 16 π 15 4 sin ⁡ 2 π 15 = sin ⁡ 16 π 15 cos ⁡ 16 π 15 8 sin ⁡ 2 π 15 = sin ⁡ 32 π 15 16 sin ⁡ 2 π 15 = sin ⁡ ( 2 π + 2 π 15 ) 16 sin ⁡ 2 π 15 = sin ⁡ 2 π 15 16 sin ⁡ 2 π 15 = 1 16 = 0.0625 \begin{aligned} P & = \cos \frac {2\pi}{15} \cos \frac {4\pi}{15} \cos \frac {8\pi}{15} \cos \frac {16\pi}{15} \\ & = \frac {{\color{#3D99F6} \sin \frac {2\pi}{15}}\cos \frac {2\pi}{15} \cos \frac {4\pi}{15} \cos \frac {8\pi}{15} \cos \frac {16\pi}{15}}{\color{#3D99F6} \sin \frac {2\pi}{15}} \\ & = \frac {{\color{#3D99F6} \sin \frac {4\pi}{15}}\cos \frac {4\pi}{15} \cos \frac {8\pi}{15} \cos \frac {16\pi}{15}}{{\color{#3D99F6}2} \sin \frac {2\pi}{15}} \\ & = \frac {{\color{#3D99F6} \sin \frac {8\pi}{15}} \cos \frac {8\pi}{15} \cos \frac {16\pi}{15}}{{\color{#3D99F6}4} \sin \frac {2\pi}{15}} \\ & = \frac {{\color{#3D99F6} \sin \frac {16\pi}{15}}\cos \frac {16\pi}{15}}{{\color{#3D99F6}8} \sin \frac {2\pi}{15}} \\ & = \frac {\color{#3D99F6}\sin \frac {32\pi}{15}}{{\color{#3D99F6}16} \sin \frac {2\pi}{15}} \\ & = \frac {\color{#3D99F6}\sin \left(2\pi + \frac {2\pi}{15}\right)}{16 \sin \frac {2\pi}{15}} \\ & = \frac {\color{#3D99F6}\sin \frac {2\pi}{15}}{16 \sin \frac {2\pi}{15}} \\ & = \frac 1{16} = \boxed{0.0625} \end{aligned}

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