Trigo + Binomial!

Probability Level 4

1 + 1 3 + 1 3 ⋅ 3 6 + 1 3 ⋅ 3 6 ⋅ 5 9 + 1 3 ⋅ 3 6 ⋅ 5 9 ⋅ 7 12 + ⋯ = 2 cos ⁡ θ 1+\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{3}{6}+\dfrac{1}{3}\cdot\dfrac{3}{6}\cdot\dfrac{5}{9}+\dfrac{1}{3}\cdot\dfrac{3}{6}\cdot\dfrac{5}{9}\cdot\dfrac{7}{12}+\cdots=2\cos\theta

Find the value of θ \theta in degrees if 0 ≤ θ ≤ π 0 \le \theta \le π .


The answer is 30.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

Chew-Seong Cheong
Apr 25, 2018

Considering the binomial expansion of 1 1 − x \dfrac 1{\sqrt{1-x}} :

( 1 − x ) − 1 2 = 1 + 1 2 ⋅ x 1 ! + 1 2 ⋅ 3 2 ⋅ x 2 2 ! + 1 2 ⋅ 3 2 ⋅ 5 2 ⋅ x 3 3 ! + 1 2 ⋅ 3 2 ⋅ 5 2 ⋅ 7 2 ⋅ 4 3 4 ! + ⋯ Putting x = 2 3 ⟹ 1 1 − 2 3 = 1 + 1 3 ⋅ 1 ! + 1 ⋅ 3 3 2 ⋅ 2 ! + 1 ⋅ 3 ⋅ 5 3 3 ⋅ 3 ! + 1 ⋅ 3 ⋅ 5 ⋅ 7 3 4 ⋅ 4 ! + ⋯ 3 = 1 + 1 3 + 1 3 ⋅ 3 6 + 1 3 ⋅ 3 6 ⋅ 5 9 + 1 3 ⋅ 3 6 ⋅ 5 9 ⋅ 7 12 + ⋯ \begin{aligned} (1-x)^{-\frac 12} & = 1 + \frac 12\cdot \frac x{1!} + \frac 12 \cdot \frac 32 \cdot \frac {x^2}{2!} + \frac 12 \cdot \frac 32 \cdot \frac 52 \cdot \frac {x^3}{3!} + \frac 12 \cdot \frac 32 \cdot \frac 52 \cdot \frac 72 \cdot \frac {4^3}{4!} + \cdots & \small \color{#3D99F6} \text{Putting }x = \frac 23 \\ \implies \frac 1{\sqrt{1-\frac 23}} & = 1 + \frac 1{3\cdot 1!} + \frac {1\cdot 3}{3^2\cdot 2!} + \frac {1\cdot 3 \cdot 5}{3^3\cdot 3!} + \frac {1\cdot 3 \cdot 5 \cdot 7}{3^4\cdot 4!} + \cdots \\ \sqrt 3 & = 1 + \frac 13 + \frac 13 \cdot \frac 36 + \frac 13 \cdot \frac 36 \cdot \frac 59 + \frac 13 \cdot \frac 36 \cdot \frac 59 \cdot \frac 7{12} + \cdots \end{aligned}

Therefore 2 cos ⁡ θ = 3 2\cos \theta = \sqrt 3 ⟹ cos ⁡ θ = 3 2 \implies \cos \theta = \dfrac {\sqrt 3}2 ⟹ θ = 30 ∘ \implies \theta = \boxed{30}^\circ for 0 ∘ ≤ θ ≤ 18 0 ∘ 0^\circ \le \theta \le 180^\circ .

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Did same here.A very easy and oral question if one knows basic binomial expansions.Further is just a formality.

D K - 2 years, 10 months ago
Vitor Juiz
Apr 25, 2018

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution

LUCAS MACHADO - 3 years, 1 month ago

Actually when solving for a general case you should not write it as the "C" notation ( or the binomial coefficient notation) as it is valid if n is a positive integer . Although it looks the same after evaluating it is not proper. You should write nC2 simply as n(n-1)/2 ( as this is derived from differentiation and not pascals triangle as original binomial theorem is ) .
You can use the "C" notation when we are dealing with n as a positive integer.

Arghyadeep Chatterjee - 3 years, 1 month ago

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Thanks. I'll do that in my future solutions.

Vitor Juiz - 3 years, 1 month ago

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