Sine and cosine sound the same?

Geometry Level 3

In school, Jimmy was told to calculate the value of the series sin ⁡ x + sin ⁡ 2 x + sin ⁡ 3 x + sin ⁡ 4 x + ⋯ \sin x + \sin^2 x + \sin^3x + \sin^4x + \cdots for some known x x . But Jimmy misheard, and he instead calculated the value of the series cos ⁡ x + cos ⁡ 2 x + cos ⁡ 3 x + cos ⁡ 4 x + ⋯ . \cos x + \cos^2 x + \cos^3x + \cos^4x + \cdots . Surprisingly, Jimmy got the correct answer!

Assuming that he did his work correctly and that his answer was a negative number, what was the answer?

3 − 10 3 - \sqrt{10} 2 − 5 2 - \sqrt5 1 − 2 1 - \sqrt2 4 − 17 4 - \sqrt{17}

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Pi Han Goh by Pi Han Goh , 30, Malaysia

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2 solutions

Michael Huang
Dec 3, 2016

Consider (see Note ) ∑ n = 1 ∞ x n = − x x − 1 \sum\limits_{n = 1}^{\infty} x^n = -\dfrac{x}{x - 1} Since Jimmy obtains the same results from both cos ⁡ \cos and sin ⁡ \sin series, we want to find θ \theta , such that ∑ n = 1 ∞ ( cos ⁡ ( θ ) ) n = ∑ n = 1 ∞ ( sin ⁡ ( θ ) ) n \sum\limits_{n = 1}^{\infty} \left(\cos\left(\theta\right)\right)^n = \sum\limits_{n = 1}^{\infty} \left(\sin\left(\theta\right)\right)^n which is also − cos ⁡ ( θ ) cos ⁡ ( θ ) − 1 = − sin ⁡ ( θ ) sin ⁡ ( θ ) − 1 cos ⁡ ( θ ) cos ⁡ ( θ ) − 1 = sin ⁡ ( θ ) sin ⁡ ( θ ) − 1 \begin{array}{rl} -\dfrac{\cos\left(\theta\right)}{\cos\left(\theta\right) - 1} &= -\dfrac{\sin\left(\theta\right)}{\sin\left(\theta\right) - 1}\\ \dfrac{\cos\left(\theta\right)}{\cos\left(\theta\right) - 1} &= \dfrac{\sin\left(\theta\right)}{\sin\left(\theta\right) - 1} \end{array} Solving for θ \theta , ( sin ⁡ ( θ ) − 1 ) ( cos ⁡ ( θ ) − 1 ) cos ⁡ ( θ ) ⋅ cos ⁡ ( θ ) cos ⁡ ( θ ) − 1 = sin ⁡ ( θ ) sin ⁡ ( θ ) − 1 ⋅ ( sin ⁡ ( θ ) − 1 ) ( cos ⁡ ( θ ) − 1 ) cos ⁡ ( θ ) sin ⁡ ( θ ) − 1 = sin ⁡ ( θ ) cos ⁡ ( θ ) ( cos ⁡ ( θ ) − 1 ) sin ⁡ ( θ ) − 1 = sin ⁡ ( θ ) − sin ⁡ ( θ ) cos ⁡ ( θ ) − 1 = − tan ⁡ ( θ ) ⟹ θ = π k − 3 π 4 k ∈ Z \begin{array}{rl} \dfrac{\left(\sin(\theta) - 1\right)\left(\cos(\theta) - 1\right)}{\cos(\theta)} \cdot \dfrac{\cos\left(\theta\right)}{\cos\left(\theta\right) - 1} &= \dfrac{\sin\left(\theta\right)}{\sin\left(\theta\right) - 1} \cdot \dfrac{\left(\sin(\theta) - 1\right)\left(\cos(\theta) - 1\right)}{\cos(\theta)}\\ \sin(\theta) - 1 &= \dfrac{\sin(\theta)}{\cos(\theta)}\left(\cos(\theta) - 1\right)\\ \sin(\theta) - 1 &= \sin(\theta) - \dfrac{\sin(\theta)}{\cos(\theta)}\\ -1 &= -\tan(\theta)\\ \Longrightarrow \theta &= \pi k - \dfrac{3\pi}{4} \quad k \in \mathbb{Z} \end{array} But since Jimmy computed the negative value of the series, this concludes 2 π k − 3 π 4 2\pi k - \dfrac{3\pi}{4} to be the solutions for the problem. Thus, cos ⁡ ( 2 π k − 3 π 4 ) cos ⁡ ( 2 π k − 3 π 4 ) − 1 = sin ⁡ ( 2 π k − 3 π 4 ) sin ⁡ ( 2 π k − 3 π 4 ) − 1 = 1 − 2 \dfrac{\cos\left(2\pi k -\dfrac{3\pi}{4}\right)}{\cos\left(2\pi k -\dfrac{3\pi}{4}\right) - 1} = \dfrac{\sin\left(2\pi k -\dfrac{3\pi}{4}\right)}{\sin\left(2\pi k -\dfrac{3\pi}{4}\right) - 1} = \boxed{1 - \sqrt{2}}

Note

To prove that ∑ n = 1 ∞ x n = − x x − 1 \sum\limits_{n = 1}^{\infty} x^n = -\dfrac{x}{x - 1} apply the following Taylor series

∑ n = 0 ∞ x n = 1 1 − x \sum\limits_{n = 0}^{\infty} x^n = \dfrac{1}{1 - x}

which gives ∑ n = 1 ∞ x n = ( ∑ n = 0 ∞ x n ) − 1 = 1 1 − x − 1 = 1 − ( 1 − x ) 1 − x = x 1 − x = − x x − 1 \begin{array}{rl} \sum\limits_{n = 1}^{\infty} x^n &= \left(\sum\limits_{n = 0}^{\infty} x^n\right) - 1\\ &= \dfrac{1}{1 - x} - 1\\ &= \dfrac{1 - (1 - x)}{1 - x}\\ &= \dfrac{x}{1 - x}\\ &= \dfrac{-x}{x - 1} \end{array}

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This is a very detailed and neat answer. Thank you!!

On the other hand, you can just prove that ∑ n = 1 ∞ x n = − x x − 1 \displaystyle \sum_{n=1}^\infty x^n = -\dfrac x{x-1} using the geometric progression sum formula.

Pi Han Goh - 4 years, 6 months ago

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Certainly!

Michael Huang - 4 years, 6 months ago

The answer here does not show that the solution is unique. See comments below.

We note that sin ⁡ x = cos ⁡ x \sin x = \cos x , only when sin ⁡ x = cos ⁡ x = ± 1 2 \sin x = \cos x = \pm \dfrac 1{\sqrt 2} . Jimmy must be calculating:

∑ k = 1 ∞ ( − 1 2 ) k = − 1 2 ( 1 1 + 1 2 ) = − 1 2 ( 2 2 + 1 ) = − 1 2 + 1 = 1 − 2 < 0 \begin{aligned} \sum_{k=1}^\infty \left(-\frac 1 {\sqrt 2} \right)^k & = -\frac 1 {\sqrt 2} \left(\frac 1 {1+\frac 1 {\sqrt 2}} \right) \\ & = -\frac 1 {\sqrt 2} \left(\frac {\sqrt 2 }{\sqrt 2+1 } \right) \\ & = -\frac 1{\sqrt 2+1 } \\ & = \boxed {1-\sqrt 2} <0 \end{aligned}

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If a 1 + a 2 + a 3 + a 4 + ⋯ = b 1 + b 2 + b 3 + b 4 + ⋯ a_1 + a_2 + a_3 + a_4 + \cdots = b_1 + b_2 + b_3 + b_4 + \cdots , then it is possible that a_1 = b_1, a_2 = b_2 , a_3 = b_3 , \ldots . But does that mean that it is the only possible solution?

That is, how do you know that a 1 = b 1 + b 2 , a 2 = b 3 + b 4 , a 3 = b 5 + b 6 , … a_1 = b_1 + b_2 , a_2 = b_3 + b_4 , a_3 = b_5 + b_6 , \ldots can't be true?

Pi Han Goh - 4 years, 6 months ago

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Very good point! a n = b n a_n = b_n does not have to be that way for the sum to be equivalent. For instance, if a n = ( − 1 ) n a_n = (-1)^{n} and b n = ( − 1 ) n + 1 b_n = (-1)^{n + 1} , then a n ≠ b n a_n \neq b_n .

I understand the point that sin ⁡ = cos ⁡ \sin = \cos . He is actually right about this since for distinct integers m , n m,n cos ⁡ m ≠ c o s n \cos^m \neq cos^n and sin ⁡ m ≠ sin ⁡ n \sin^m \neq \sin^n However, I believe that Pi Han's point shows that setting a n = b n a_n = b_n is not that obvious since there may be other terms that are also equivalent.

Michael Huang - 4 years, 6 months ago

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