Trigonometry! #46

Geometry Level 3

Simplify 16 cos ⁡ ( 2 π 15 ) cos ⁡ ( 4 π 15 ) cos ⁡ ( 8 π 15 ) cos ⁡ ( 16 π 15 ) 16 \cos \left (\frac {2\pi}{15} \right) \cos \left (\frac {4\pi}{15} \right) \cos \left (\frac {8\pi}{15} \right) \cos \left (\frac {16\pi}{15} \right)

This problem is part of the set Trigonometry .


The answer is 1.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

Jared Low
Jan 30, 2015

16 cos ⁡ ( 2 π 15 ) cos ⁡ ( 4 π 15 ) cos ⁡ ( 8 π 15 ) cos ⁡ ( 16 π 15 ) 16\cos(\frac{2\pi}{15})\cos(\frac{4\pi}{15})\cos(\frac{8\pi}{15})\cos(\frac{16\pi}{15})

= 16 sin ⁡ ( 2 π 15 ) cos ⁡ ( 2 π 15 ) cos ⁡ ( 4 π 15 ) cos ⁡ ( 8 π 15 ) cos ⁡ ( 16 π 15 ) ÷ sin ⁡ ( 2 π 15 ) =16\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})\div \sin(\frac{2\pi}{15})

= 8 sin ⁡ ( 4 π 15 ) cos ⁡ ( 4 π 15 ) cos ⁡ ( 8 π 15 ) cos ⁡ ( 16 π 15 ) ÷ sin ⁡ ( 2 π 15 ) =8\sin(\frac{4\pi}{15})\cos(\frac{4\pi}{15})\cos(\frac{8\pi}{15})\cos(\frac{16\pi}{15})\div \sin(\frac{2\pi}{15})

= 4 sin ⁡ ( 8 π 15 ) cos ⁡ ( 8 π 15 ) cos ⁡ ( 16 π 15 ) ÷ sin ⁡ ( 2 π 15 ) =4\sin(\frac{8\pi}{15})\cos(\frac{8\pi}{15})\cos(\frac{16\pi}{15})\div \sin(\frac{2\pi}{15})

= 2 sin ⁡ ( 16 π 15 ) cos ⁡ ( 16 π 15 ) ÷ sin ⁡ ( 2 π 15 ) =2\sin(\frac{16\pi}{15})\cos(\frac{16\pi}{15})\div \sin(\frac{2\pi}{15})

= sin ⁡ ( 32 π 15 ) ÷ sin ⁡ ( 2 π 15 ) =\sin(\frac{32\pi}{15})\div \sin(\frac{2\pi}{15})

= sin ⁡ ( 2 π 15 ) ÷ sin ⁡ ( 2 π 15 ) = 1 =\sin(\frac{2\pi}{15})\div \sin(\frac{2\pi}{15})=\boxed{1}

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