Product of cosines (or sines)

Geometry Level 4

If θ = π 13 \theta = \dfrac \pi{13} , determine the value of the reciprocal of

cos ⁡ θ cos ⁡ 2 θ cos ⁡ 3 θ cos ⁡ 4 θ cos ⁡ 5 θ cos ⁡ 6 θ \cosθ\cos2θ\cos3θ\cos4θ\cos5θ\cos6θ .


The answer is 64.

Shubhrajit Sadhukhan by Shubhrajit Sadhukhan , 16, India

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2 solutions

Kushal Dey
Apr 20, 2021

Okay, we have 13θ=π, and we have to compute the given expression. Note that, since 13 is a prime number, it makes this process possible.
Let P=cos(θ)cos(2θ)cos(3θ)...cos(6θ).
Since, 13θ=π,
cos(5θ)=cos(5θ-13θ)=-cos(-8θ)=-cos(8θ).
Then, cos(3θ)=-cos(3θ+13θ)=-cos(16θ).
And also, cos(6θ)=cos(6θ+26θ)=cos(32θ).
Thus P=cos(θ)cos(2θ)...cos(32θ). This expression is actually a "telescopic product". For those who are not familiar with the concept, consider this,
sin(2θ)/(2sin(θ))=cos(θ),
sin(4θ)/(2sin(2θ))=cos(2θ), sin(8θ)/(2sin(4θ))=cos(4θ),
....
sin(64θ)/(2sin(32θ))=cos(32θ).
Multiply all these equations and you will get, sin(64θ)/(64sin(θ))=P. But again we write sin(64θ)=sin(65θ-64θ)=sin(θ). Thus we have, P=1/64 => 1/P=64.


3 Helpful 2 Interesting 2 Brilliant 0 Confused

Note that, previously we had the expression as product of cosines of angles in AP, which we converted into a product of cosines of angles in GP. It was only possible because 13 is relatively prime to 2, due to which we know that any integer power of 2, will have remainder that is also relatively prime to 13 when divided by 13. Since 13 was a prime to begin with, it was possible to convert all numbers smaller than it as a power of 2.

Kushal Dey - 1 month, 3 weeks ago

I also want to add 2 more solutions, but how do i do that? After writing this solution, i am not getting any options to write another.

Kushal Dey - 1 month, 3 weeks ago

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Nice solution! You only get one "space" to write a solution on here, but you can either add extra solutions in the comments, or you can divide up the main solution box by typing a single underscore character on a line, which gives you this:

...so you can have a different section. It'll make it easier to read if you write the solutions in LaTeX (I see you did for the question).

Chris Lewis - 1 month, 3 weeks ago

Thank you Chris, that was helpful.

Kushal Dey - 1 month, 2 weeks ago
Chew-Seong Cheong
Apr 22, 2021

x = 1 cos ⁡ θ cos ⁡ 2 θ cos ⁡ 3 θ cos ⁡ 4 θ cos ⁡ 5 θ cos ⁡ 6 θ Since θ = π 13 = 1 cos ⁡ π 13 cos ⁡ 2 π 13 cos ⁡ 3 π 13 cos ⁡ 4 π 13 cos ⁡ 5 π 13 cos ⁡ 6 π 13 Note that cos ⁡ ( π − ϕ ) = − cos ⁡ ϕ = 1 cos ⁡ π 13 cos ⁡ 2 π 13 cos ⁡ 4 π 13 ( − cos ⁡ 8 π 13 ) cos ⁡ 3 π 13 cos ⁡ 6 π 13 = − sin ⁡ π 13 sin ⁡ π 13 cos ⁡ π 13 cos ⁡ 2 π 13 cos ⁡ 4 π 13 cos ⁡ 8 π 13 cos ⁡ 3 π 13 cos ⁡ 6 π 13 = − 2 sin ⁡ π 13 sin ⁡ 2 π 13 cos ⁡ 2 π 13 cos ⁡ 4 π 13 cos ⁡ 8 π 13 cos ⁡ 3 π 13 cos ⁡ 6 π 13 = − 4 sin ⁡ π 13 sin ⁡ 4 π 13 cos ⁡ 4 π 13 cos ⁡ 8 π 13 cos ⁡ 3 π 13 cos ⁡ 6 π 13 = − 8 sin ⁡ π 13 sin ⁡ 8 π 13 cos ⁡ 8 π 13 cos ⁡ 3 π 13 cos ⁡ 6 π 13 = − 16 sin ⁡ π 13 sin ⁡ 16 π 13 cos ⁡ 3 π 13 cos ⁡ 6 π 13 and sin ⁡ ( π − ϕ ) = sin ⁡ ϕ = − 16 sin ⁡ π 13 − sin ⁡ 3 π 13 cos ⁡ 3 π 13 cos ⁡ 6 π 13 = 32 sin ⁡ π 13 sin ⁡ 6 π 13 cos ⁡ 6 π 13 = 64 sin ⁡ π 13 sin ⁡ 12 π 13 = 64 sin ⁡ π 13 sin ⁡ π 13 = 64 \begin{aligned} x & = \frac 1{\cos \theta \cos 2\theta \cos 3\theta \cos 4\theta \cos 5\theta \cos 6\theta} & \small \blue{\text{Since }\theta = \frac \pi{13}} \\ & = \frac 1{\cos \frac \pi{13} \cos \frac {2\pi}{13} \cos \frac {3\pi}{13} \cos \frac {4\pi}{13} \blue{\cos \frac {5\pi}{13}} \cos \frac {6\pi}{13}} & \small \blue{\text{Note that }\cos (\pi - \phi) = - \cos \phi} \\ & = \frac 1{\cos \frac \pi{13} \cos \frac {2\pi}{13} \cos \frac {4\pi}{13} \blue{\left(-\cos \frac {8\pi}{13}\right)} \cos \frac {3\pi}{13} \cos \frac {6\pi}{13}} \\ & = \frac {-\blue{\sin \frac \pi{13}}}{\blue{\sin \frac \pi{13}} \cos \frac \pi{13} \cos \frac {2\pi}{13} \cos \frac {4\pi}{13} \cos \frac {8\pi}{13} \cos \frac {3\pi}{13} \cos \frac {6\pi}{13}} \\ & = \frac {-\blue 2 \sin \frac \pi{13}}{\blue{\sin \frac {2\pi}{13}} \cos \frac {2\pi}{13} \cos \frac {4\pi}{13} \cos \frac {8\pi}{13} \cos \frac {3\pi}{13} \cos \frac {6\pi}{13}} \\ & = \frac {-\blue 4 \sin \frac \pi{13}}{\blue{\sin \frac {4\pi}{13}} \cos \frac {4\pi}{13} \cos \frac {8\pi}{13} \cos \frac {3\pi}{13} \cos \frac {6\pi}{13}} \\ & = \frac {-\blue 8 \sin \frac \pi{13}}{\blue{\sin \frac {8\pi}{13}} \cos \frac {8\pi}{13} \cos \frac {3\pi}{13} \cos \frac {6\pi}{13}} \\ & = \frac {-\blue {16} \sin \frac \pi{13}}{\blue{\sin \frac {16\pi}{13}} \cos \frac {3\pi}{13} \cos \frac {6\pi}{13}} & \small \blue{\text{and }\sin (\pi - \phi) = \sin \phi} \\ & = \frac {- 16 \sin \frac \pi{13}}{\blue{-\sin \frac {3\pi}{13}} \cos \frac {3\pi}{13} \cos \frac {6\pi}{13}} \\ & = \frac {\blue{32} \sin \frac \pi{13}}{\blue{\sin \frac {6\pi}{13}} \cos \frac {6\pi}{13}} = \frac {\blue {64} \sin \frac \pi{13}}{\blue{\sin \frac {12\pi}{13}}} = \frac {64 \sin \frac \pi{13}}{\blue{\sin \frac \pi{13}}} = \boxed{64} \end{aligned}

1 Helpful 1 Interesting 1 Brilliant 1 Confused

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