The power of trig identities!

Geometry Level 3

Find the value of cos ⁡ π 15 cos ⁡ 2 π 15 cos ⁡ 3 π 15 cos ⁡ 4 π 15 cos ⁡ 5 π 15 cos ⁡ 6 π 15 cos ⁡ 7 π 15 \cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}

1 64 \frac{1}{64} 1 512 \frac{1}{512} 1 128 \frac{1}{128} 1 32 \frac{1}{32} 1 256 \frac{1}{256}

ChengYiin Ong by ChengYiin Ong , 17, Malaysia

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2 solutions

ChengYiin Ong
Mar 20, 2020

We will first calculate three values: cos ⁡ π 15 cos ⁡ 2 π 15 cos ⁡ 4 π 15 cos ⁡ 7 π 15 \cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{7\pi}{15} cos ⁡ 3 π 15 cos ⁡ 6 π 15 \cos \frac{3\pi}{15} \cos \frac{6\pi}{15} cos ⁡ 5 π 15 \cos \frac{5\pi}{15} The first value is given by:

cos ⁡ π 15 cos ⁡ 2 π 15 cos ⁡ 4 π 15 cos ⁡ 7 π 15 \cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{7\pi}{15}

= 16 cos ⁡ π 15 sin ⁡ π 15 cos ⁡ 2 π 15 cos ⁡ 4 π 15 cos ⁡ 7 π 15 16 sin ⁡ π 15 =\frac{16\cos \frac{\pi}{15}\sin \frac{\pi}{15}\cos \frac{2\pi}{15}\cos \frac{4\pi}{15}\cos \frac{7\pi}{15}}{16\sin \frac{\pi}{15}} = 8 sin ⁡ 2 π 15 cos ⁡ 2 π 15 cos ⁡ 4 π 15 cos ⁡ 7 π 15 16 sin ⁡ π 15 =\frac{8\sin \frac{2\pi}{15}\cos \frac{2\pi}{15}\cos \frac{4\pi}{15}\cos \frac{7\pi}{15}}{16\sin \frac{\pi}{15}} = 4 sin ⁡ 4 π 15 cos ⁡ 4 π 15 cos ⁡ 7 π 15 16 sin ⁡ π 15 =\frac{4\sin \frac{4\pi}{15}\cos \frac{4\pi}{15}\cos \frac{7\pi}{15}}{16\sin \frac{\pi}{15}} = 2 sin ⁡ 8 π 15 cos ⁡ 7 π 15 16 sin ⁡ π 15 =\frac{2\sin \frac{8\pi}{15}\cos \frac{7\pi}{15}}{16\sin \frac{\pi}{15}} = 2 sin ⁡ 7 π 15 cos ⁡ 7 π 15 16 sin ⁡ π 15 =\frac{2\sin \frac{7\pi}{15}\cos \frac{7\pi}{15}}{16\sin \frac{\pi}{15}} = sin ⁡ 14 π 15 16 sin ⁡ π 15 =\frac{\sin \frac{14\pi}{15}}{16\sin \frac{\pi}{15}} = sin ⁡ π 15 16 sin ⁡ π 15 =\frac{\sin \frac{\pi}{15}}{16\sin \frac{\pi}{15}} = 1 16 =\frac{1}{16}

The second value is given by: cos ⁡ 3 π 15 cos ⁡ 6 π 15 \cos \frac{3\pi}{15} \cos \frac{6\pi}{15} = 4 sin ⁡ 3 π 15 cos ⁡ 3 π 15 cos ⁡ 6 π 15 4 sin ⁡ 3 π 15 =\frac{4\sin \frac{3\pi}{15} \cos \frac{3\pi}{15} \cos \frac{6\pi}{15}}{4\sin \frac{3\pi}{15}} = 2 sin ⁡ 6 π 15 cos ⁡ 6 π 15 4 sin ⁡ 3 π 15 =\frac{2\sin \frac{6\pi}{15} \cos \frac{6\pi}{15}}{4\sin \frac{3\pi}{15}} = sin ⁡ 12 π 15 4 sin ⁡ 3 π 15 =\frac{\sin \frac{12\pi}{15}}{4\sin \frac{3\pi}{15}} = sin ⁡ 3 π 15 4 sin ⁡ 3 π 15 =\frac{\sin \frac{3\pi}{15}}{4\sin \frac{3\pi}{15}} = 1 4 =\frac{1}{4}

The third value is given by: cos ⁡ 5 π 15 \cos \frac{5\pi}{15} = 1 2 =\frac{1}{2}

Thus, the desired value:

cos ⁡ π 15 cos ⁡ 2 π 15 cos ⁡ 3 π 15 cos ⁡ 4 π 15 cos ⁡ 5 π 15 cos ⁡ 6 π 15 cos ⁡ 7 π 15 \cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15} = ( 1 16 ) ( 1 4 ) ( 1 2 ) =(\frac{1}{16})(\frac{1}{4})(\frac{1}{2}) = 1 128 =\frac{1}{128}

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Chew-Seong Cheong
Mar 21, 2020

P = cos ⁡ π 15 cos ⁡ 2 π 15 cos ⁡ 3 π 15 cos ⁡ 4 π 15 cos ⁡ 5 π 15 cos ⁡ 6 π 15 cos ⁡ 7 π 15 = cos ⁡ π 15 cos ⁡ 2 π 15 cos ⁡ 4 π 15 cos ⁡ 7 π 15 × cos ⁡ 3 π 15 cos ⁡ 6 π 15 × cos ⁡ 5 π 15 Note that cos ⁡ ( π − θ ) = − cos ⁡ θ = cos ⁡ π 15 cos ⁡ 2 π 15 cos ⁡ 4 π 15 ( − cos ⁡ 8 π 15 ) × cos ⁡ π 5 cos ⁡ 2 π 5 × cos ⁡ π 3 and cos ⁡ A cos ⁡ B = 1 2 ( cos ⁡ ( A − B ) + cos ⁡ ( A + B ) ) = − sin ⁡ π 15 cos ⁡ π 15 cos ⁡ 2 π 15 cos ⁡ 4 π 15 cos ⁡ 8 π 15 sin ⁡ π 15 × 1 2 ( cos ⁡ π 5 + cos ⁡ 3 π 5 ) × 1 2 See proof: ∑ k = 0 n − 1 cos ⁡ ( 2 k + 1 2 n + 1 π ) = 1 2 = − sin ⁡ 2 π 15 cos ⁡ 2 π 15 cos ⁡ 4 π 15 cos ⁡ 8 π 15 2 sin ⁡ π 15 × 1 2 ( 1 2 ) × 1 2 = − sin ⁡ 4 π 15 cos ⁡ 4 π 15 cos ⁡ 8 π 15 4 sin ⁡ π 15 × 1 8 = − sin ⁡ 8 π 15 cos ⁡ 8 π 15 8 sin ⁡ π 15 × 1 8 = − sin ⁡ 16 π 15 16 sin ⁡ π 15 × 1 8 = − − sin ⁡ π 15 16 sin ⁡ π 15 × 1 8 = 1 128 \begin{aligned} P & = \cos \frac \pi{15} \cos \frac {2\pi}{15} \cos \frac {3\pi}{15} \cos \frac {4\pi}{15} \cos \frac {5\pi}{15} \cos \frac {6\pi}{15} \cos \frac {7\pi}{15} \\ & = \cos \frac \pi{15} \cos \frac {2\pi}{15} \cos \frac {4\pi}{15} \blue{\cos \frac {7\pi}{15}} \times \cos \frac {3\pi}{15} \cos \frac {6\pi}{15} \times \cos \frac {5\pi}{15} & \small \blue{\text{Note that }\cos (\pi - \theta) = - \cos \theta} \\ & = \small \cos \frac \pi{15} \cos \frac {2\pi}{15} \cos \frac {4\pi}{15} \blue{\left(-\cos \frac {8\pi}{15}\right)} \times \red{\cos \frac \pi 5 \cos \frac {2\pi}5} \times \cos \frac \pi 3 & \small \red{\text{and }\cos A \cos B = \frac 12 (\cos(A-B) + \cos (A+B))} \\ & = \small - \frac {\blue{\sin \frac \pi{15}} \cos \frac \pi{15} \cos \frac {2\pi}{15} \cos \frac {4\pi}{15} \cos \frac {8\pi}{15}}{\blue{\sin \frac \pi{15}}} \times \red{\frac 12\left(\cos \frac \pi 5 + \cos \frac {3\pi}5 \right)} \times \frac 12 & \small \red{\text{See proof: }\sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) = \frac 12} \\ & = \small - \frac {\sin \frac {2\pi}{15} \cos \frac {2\pi}{15} \cos \frac {4\pi}{15} \cos \frac {8\pi}{15}}{2\sin \frac \pi{15}} \times \red{\frac 12\left(\frac 12 \right)} \times \frac 12 \\ & = \small - \frac {\sin \frac {4\pi}{15} \cos \frac {4\pi}{15} \cos \frac {8\pi}{15}}{4\sin \frac \pi{15}} \times \frac 18 \\ & = \small - \frac {\sin \frac {8\pi}{15} \cos \frac {8\pi}{15}}{8\sin \frac \pi{15}} \times \frac 18 \\ & = \small - \frac {\blue{\sin \frac {16\pi}{15}}}{16\sin \frac \pi{15}} \times \frac 18 = - \frac {\blue{-\sin \frac \pi{15}}}{16\sin \frac \pi{15}} \times \frac 18 \\ & = \boxed{\frac 1{128}} \end{aligned}

Reference: Proof for ∑ k = 0 n − 1 cos ⁡ ( 2 k + 1 2 n + 1 π ) = 1 2 \displaystyle \sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) = \frac 12

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