200 Followers Problem: Trigonometric sum from the future

Geometry Level 5

S = ∑ k = 1 2020 ( 1 sin ⁡ ( 2 k − 1 4040 π + θ ) ⋅ sin ⁡ ( 2 k − 1 4040 π − θ ) ) \large \text{S} = \sum_{k=1}^{2020} \left(\dfrac{1}{\sin \left(\frac{2k-1}{4040}\pi + \theta \right) \cdot \sin \left(\frac{2k-1}{4040} \pi - \theta \right)}\right)

where 0 < θ < π 2 0< \theta < \dfrac{\pi}{2}

If S \text{S} can be represented as:

A ⋅ ( tan ⁡ ( B θ ) sin ⁡ ( C θ ) ) \large A \cdot \left( \dfrac{\tan (B \theta)}{\sin (C \theta)} \right)

where A A , B B and C C are positive integers, evaluate A + B + C A+B+C .

Original


The answer is 6062.

Ishan Singh by Ishan Singh , 22, India

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1 solution

Ishan Singh
Oct 4, 2015

Consider the general summation,

S = ∑ k = 1 n ( 1 sin ⁡ ( 2 k − 1 2 n π + θ ) ⋅ sin ⁡ ( 2 k − 1 2 n π − θ ) ) \displaystyle \text{S} = \sum_{k=1}^{n} \left(\dfrac{1}{\sin \left(\dfrac{2k-1}{2n}\pi + \theta \right) \cdot \sin \left(\dfrac{2k-1}{2n}\pi - \theta \right)}\right)

= ∑ k = 1 n ( 1 cos ⁡ 2 θ − cos ⁡ 2 ( 2 k − 1 2 n π ) ) = \displaystyle \sum_{k=1}^{n} \left(\dfrac{1}{\cos^2 \theta - \cos^2 \left(\dfrac{2k-1}{2n}\pi\right)}\right)

= ∑ k = 1 n 1 2 cos ⁡ θ ( 1 cos ⁡ θ − cos ⁡ ( 2 k − 1 2 n π ) + 1 cos ⁡ θ + cos ⁡ ( 2 k − 1 2 n π ) ) = \displaystyle \sum_{k=1}^{n} \dfrac{1}{2\cos \theta} \left(\dfrac{1}{\cos \theta - \cos \left(\dfrac{2k-1}{2n}\pi\right)} + \dfrac{1}{\cos \theta + \cos \left(\dfrac{2k-1}{2n}\pi\right)} \right)

= ∑ k = 1 n 1 cos ⁡ θ ( 1 cos ⁡ θ − cos ⁡ ( 2 k − 1 2 n π ) ) ( ∵ ∑ k = 1 n f ( k ) = ∑ k = 1 n f ( n + 1 − k ) ) = \displaystyle \sum_{k=1}^{n} \dfrac{1}{\cos \theta} \left(\dfrac{1}{\cos \theta - \cos \left(\dfrac{2k-1}{2n}\pi\right)}\right) \quad \left( \because \sum_{k=1}^{n} f(k) = \sum_{k=1}^{n} f(n+1-k) \right)

Now,

T n ( x ) = 2 n − 1 ∏ k = 1 n ( x − cos ⁡ ( 2 k − 1 2 n π ) ) \displaystyle T_{n} (x) = 2^{n-1} \prod_{k=1}^{n} \left(x - \cos \left(\dfrac{2k-1}{2n}\pi\right) \right)

where T n ( x ) T_{n}(x) is Chebyshev Polynomial of the first kind.

Taking logarithm and differentiating w.r.t. x x , we have,

∑ k = 1 n ( 1 x − cos ⁡ ( 2 k − 1 2 n π ) ) = T n ′ ( x ) T n ( x ) \displaystyle \sum_{k=1}^{n} \left(\dfrac{1}{x - \cos \left(\dfrac{2k-1}{2n}\pi\right)}\right) = \dfrac{T_{n}'(x)}{T_{n}(x)}

where T n ′ ( x ) T_{n}'(x) denotes the derivative of T n ( x ) T_{n}(x) w.r.t. x x

⟹ ∑ k = 1 n ( 1 x − cos ⁡ ( 2 k − 1 2 n π ) ) = n U n − 1 ( x ) T n ( x ) \displaystyle \implies \sum_{k=1}^{n} \left(\dfrac{1}{x - \cos \left(\dfrac{2k-1}{2n}\pi\right)}\right) = \dfrac{nU_{n-1}(x)}{T_{n}(x)}

where U n ( x ) U_{n}(x) is Chebyshev Polynomial of the second kind.

Putting x = cos ⁡ θ x = \cos \theta and noting that U n − 1 ( cos ⁡ θ ) = sin ⁡ ( n θ ) sin ⁡ θ U_{n-1}(\cos \theta) = \dfrac{\sin (n\theta)}{\sin \theta} and T n ( cos ⁡ θ ) = cos ⁡ ( n θ ) T_{n}(\cos \theta) = \cos (n\theta) we have,

∑ k = 1 n ( 1 cos ⁡ θ − cos ⁡ ( 2 k − 1 2 n π ) ) = n tan ⁡ ( n θ ) sin ⁡ θ \displaystyle \sum_{k=1}^{n} \left(\dfrac{1}{\cos \theta - \cos \left(\dfrac{2k-1}{2n}\pi\right)}\right) = \dfrac{n \tan (n\theta)}{\sin \theta}

⟹ S = 2 n tan ⁡ ( n θ ) sin ⁡ ( 2 θ ) \displaystyle \implies \text{S} = \dfrac{2n \tan(n \theta)}{\sin (2\theta)}

Putting n = 2020 n = 2020 , we have,

A = 4040 , B = 2020 A = 4040, B = 2020 and C = 2 C = 2

⟹ A + B + C = 6062 \implies A+B+C = \boxed{6062}

9 Helpful 0 Interesting 0 Brilliant 0 Confused

Wow! Brilliant! Out of the top drawer, seriously!

I did it by simplifying 1 sin ⁡ ( 2 k − 1 2 n π + θ ) sin ⁡ ( 2 k − 1 2 n π − θ ) = 1 sin ⁡ ( 2 θ ) ( cot ⁡ ( 2 k − 1 2 n π − θ ) − cot ⁡ ( 2 k − 1 2 n π + θ ) ) \displaystyle \frac{1}{\sin\left(\frac{2k-1}{2n} \pi + \theta\right) \sin\left(\frac{2k-1}{2n} \pi - \theta\right)} = \frac{1}{\sin(2\theta)}\left(\cot\left(\frac{2k-1}{2n} \pi - \theta\right) - \cot\left(\frac{2k-1}{2n} \pi + \theta\right)\right)

and then "guessing the solution" by computations for some smaller n n that the sum of the "cots" is simply 2 n tan ⁡ ( n θ ) 2n \tan(n \theta) .

This method is really bad, I know, but I think that the answer(which has come out to be so simple) can be found from the sum easily and it has given encouragement to me to find out how.

Kartik Sharma - 5 years, 8 months ago

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