Play with Trigonometry

Geometry Level 3

Let the value of the given expression ∑ N = 0 ∞ cos ⁡ N θ 3 N \sum _{ N=0 }^{ \infty }{ \frac { \cos { N\theta } }{ { 3 }^{ N } } } = S then value of S to three decimal places is: Where cos ⁡ θ \cos { \theta } = 1 3 \frac{1}{3} and θ ϵ ( 0 , π 2 ) \theta \quad \epsilon \quad \left( 0,\frac { \pi }{ 2 } \right)

Inspiration: Lucas Chaves Meyles


The answer is 1.000.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

First of all, great problem! I am going to take Chew-Seong Cheong's approach. S = ∑ N = 0 ∞ cos ⁡ N θ 3 N = 1 2 ( ∑ N = 0 ∞ ( e i θ 3 ) N + ∑ N = 0 ∞ ( e − i θ 3 ) N ) = 1 2 ( 1 1 − 1 3 e i θ + 1 1 − 1 3 e − i θ ) = 3 6 − 2 e i θ + 3 6 − 2 e − i θ = 36 − 6 ( e i θ + e − i θ ) 40 − 12 ( e i θ + e − i θ ) = 9 − 3 cos ⁡ θ 10 − 6 cos ⁡ θ = 9 − 3 ( 1 3 ) 10 − 6 ( 1 3 ) = 8 8 = 1.000 \begin{aligned} S & = \sum_{N=0}^\infty \frac{\cos N\theta}{3^{N}} \\ & = \frac{1}{2}\bigg(\sum_{N=0}^\infty \bigg(\frac{e^{i\theta}}{3}\bigg)^{N} + \sum_{N=0}^\infty \bigg(\frac{e^{-i\theta}}{3}\bigg)^{N}\bigg) \\ & = \frac{1}{2} \bigg(\frac{1}{1-\frac{1}{3}e^{i\theta}} + \frac{1}{1- \frac{1}{3} e^{-i\theta}} \bigg) \\ & = \frac{3}{6-2e^{i\theta}} + \frac{3}{6-2 e^{-i\theta}} \\ & = \frac{36 - 6(e^{i\theta} + e^{-i\theta})}{40 - 12(e^{i\theta} + e^{-i\theta})} \\ & = \frac{9 - 3\cos \theta}{10 - 6\cos\theta} \\ & = \frac{9 - 3(\frac{1}{3})}{10 - 6(\frac{1}{3})} \\ & = \frac{8}{8} \\ & = \boxed{1.000} \end{aligned}

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