Recursive products of cosine function

Geometry Level 3

Let a = π 31 a = \dfrac \pi{31} . Calculate the value of

32 cos ⁡ ( a ) cos ⁡ ( 2 a ) cos ⁡ ( 4 a ) cos ⁡ ( 8 a ) cos ⁡ ( 16 a ) 32 \cos(a)\cos(2a)\cos(4a)\cos(8a)\cos(16a)


The answer is -1.

Srinivasa Gopal by Srinivasa Gopal , 53, India

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2 solutions

Srinivasa Gopal
Feb 8, 2020

The key idea is to multiply and divide the expression cos(a) * cos(2a) * cos(4a) * cos(8a) * cos(16a) by sin(a). Then, a cascaded use of the trigometric identity sin(2x) = 2 sin(x) cos(x) makes the product equal sin(32 a) / (32 sin(a)). But for the given value of a, 32a = pi + a, so sin(32 a) / sin(a) = -1 and so 32 * cos(a) * cos(2a) * cos(4a) * cos(8a) * cos(16a) = -1

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This was also easy....

Nikola Alfredi - 1 year, 3 months ago

P = 32 cos ⁡ π 31 cos ⁡ 2 π 31 cos ⁡ 4 π 31 cos ⁡ 8 π 31 cos ⁡ 16 π 31 = 32 × sin ⁡ π 31 cos ⁡ π 31 cos ⁡ 2 π 31 cos ⁡ 4 π 31 cos ⁡ 8 π 31 cos ⁡ 16 π 31 sin ⁡ π 31 = 32 × sin ⁡ 2 π 31 cos ⁡ 2 π 31 cos ⁡ 4 π 31 cos ⁡ 8 π 31 cos ⁡ 16 π 31 2 sin ⁡ π 31 = 32 × sin ⁡ 4 π 31 cos ⁡ 4 π 31 cos ⁡ 8 π 31 cos ⁡ 16 π 31 4 sin ⁡ π 31 = 32 × sin ⁡ 8 π 31 cos ⁡ 8 π 31 cos ⁡ 16 π 31 8 sin ⁡ π 31 = 32 × sin ⁡ 16 π 31 cos ⁡ 16 π 31 16 sin ⁡ π 31 = 32 × sin ⁡ 32 π 31 32 sin ⁡ π 31 = sin ⁡ 32 π 31 sin ⁡ π 31 = − sin ⁡ π 31 sin ⁡ π 31 = − 1 \begin{aligned} P & = 32 \cos \frac \pi{31} \cos \frac {2\pi}{31} \cos \frac {4\pi}{31} \cos \frac {8\pi}{31} \cos \frac {16\pi}{31} \\ & = 32 \times \frac {\blue{\sin \frac \pi{31}}\cos \frac \pi{31} \cos \frac {2\pi}{31} \cos \frac {4\pi}{31} \cos \frac {8\pi}{31} \cos \frac {16\pi}{31}}{\blue{\sin \frac \pi{31}}} \\ & = 32 \times \frac {\sin \frac {2\pi}{31}\cos \frac {2\pi}{31} \cos \frac {4\pi}{31} \cos \frac {8\pi}{31} \cos \frac {16\pi}{31}}{2\sin \frac \pi{31}} \\ & = 32 \times \frac {\sin \frac {4\pi}{31}\cos \frac {4\pi}{31} \cos \frac {8\pi}{31} \cos \frac {16\pi}{31}}{4\sin \frac \pi{31}} \\ & = 32 \times \frac {\sin \frac {8\pi}{31}\cos \frac {8\pi}{31} \cos \frac {16\pi}{31}}{8\sin \frac \pi{31}} \\ & = 32 \times \frac {\sin \frac {16\pi}{31}\cos \frac {16\pi}{31}}{16\sin \frac \pi{31}} \\ & = 32 \times \frac {\sin \frac {32\pi}{31}}{32\sin \frac \pi{31}} \\ & = \frac {\blue{\sin \frac {32\pi}{31}}}{\sin \frac \pi{31}} = \frac {\blue{-\sin \frac \pi{31}}}{\sin \frac \pi{31}} = \boxed{-1} \end{aligned}

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