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MathHardAdv 2019 • Paper 2

Trigonometric Ratio of Sum of Products Function and its Limit

For non-negative integers nn, define f(n)=∑k=0nsin⁡(k+1n+2π)sin⁡(k+2n+2π)∑k=0nsin⁡2(k+1n+2π)f(n) = \frac{\sum_{k=0}^{n} \sin\left(\frac{k + 1}{n + 2}\pi\right) \sin\left(\frac{k + 2}{n + 2}\pi\right)}{\sum_{k=0}^{n} \sin^2\left(\frac{k + 1}{n + 2}\pi\right)} Assuming cos⁡−1x\cos^{-1} x takes values in the principal range [0,π][0, \pi], which of the following statements is/are correct?

Options

A

f(4)=32f(4) = \frac{\sqrt{3}}{2}

Correct
B

lim⁡n→∞f(n)=12\lim_{n \to \infty} f(n) = \frac{1}{2}

C

If α=tan⁡(cos⁡−1f(6))\alpha = \tan(\cos^{-1} f(6)), then α2+2α−1=0\alpha^2 + 2\alpha - 1 = 0

Correct
D

sin⁡(7cos⁡−1f(5))=0\sin(7 \cos^{-1} f(5)) = 0

Correct

Topics & Concepts

Trigonometric FunctionsTrigonometric Ratios and IdentitiesSum of SeriesInverse Trigonometric Functions

Step-by-Step Solution

To find the value of f(n)f(n), let θ=πn+2\theta = \frac{\pi}{n+2}, which gives (n+2)θ=π(n+2)\theta = \pi.

The given function is: f(n)=∑k=0nsin⁡(k+1n+2π)sin⁡(k+2n+2π)∑k=0nsin⁡2(k+1n+2π)f(n) = \frac{\sum_{k=0}^{n} \sin\left(\frac{k + 1}{n + 2}\pi\right) \sin\left(\frac{k + 2}{n + 2}\pi\right)}{\sum_{k=0}^{n} \sin^2\left(\frac{k + 1}{n + 2}\pi\right)}

Let j=k+1j = k + 1. As kk ranges from 00 to nn, jj ranges from 11 to n+1n+1.

1. Simplification of the Numerator:

Let the numerator be NN: N=∑j=1n+1sin⁡(jθ)sin⁡((j+1)θ)N = \sum_{j=1}^{n+1} \sin(j\theta) \sin((j+1)\theta) Using the product-to-sum identity 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A \sin B = \cos(A-B) - \cos(A+B): N=12∑j=1n+1[cos⁡(θ)−cos⁡((2j+1)θ)]=n+12cos⁡θ−12∑j=1n+1cos⁡((2j+1)θ)N = \frac{1}{2} \sum_{j=1}^{n+1} \left[ \cos(\theta) - \cos((2j+1)\theta) \right] = \frac{n+1}{2}\cos\theta - \frac{1}{2} \sum_{j=1}^{n+1} \cos((2j+1)\theta)

Using the cosine series sum formula: ∑j=1n+1cos⁡((2j+1)θ)=sin⁡((n+1)⋅2θ2)sin⁡(2θ2)cos⁡(3θ+(2n+3)θ2)=sin⁡((n+1)θ)sin⁡θcos⁡((n+3)θ)\sum_{j=1}^{n+1} \cos((2j+1)\theta) = \frac{\sin\left(\frac{(n+1)\cdot 2\theta}{2}\right)}{\sin\left(\frac{2\theta}{2}\right)} \cos\left(\frac{3\theta + (2n+3)\theta}{2}\right) = \frac{\sin((n+1)\theta)}{\sin\theta} \cos((n+3)\theta)

Since (n+2)θ=π(n+2)\theta = \pi:

  • sin⁡((n+1)θ)=sin⁡(π−θ)=sin⁡θ\sin((n+1)\theta) = \sin(\pi - \theta) = \sin\theta
  • cos⁡((n+3)θ)=cos⁡(π+θ)=−cos⁡θ\cos((n+3)\theta) = \cos(\pi + \theta) = -\cos\theta

Substituting these into the sum: ∑j=1n+1cos⁡((2j+1)θ)=sin⁡θ(−cos⁡θ)sin⁡θ=−cos⁡θ\sum_{j=1}^{n+1} \cos((2j+1)\theta) = \frac{\sin\theta (-\cos\theta)}{\sin\theta} = -\cos\theta

Hence, the numerator becomes: N=n+12cos⁡θ−12(−cos⁡θ)=n+22cos⁡θN = \frac{n+1}{2}\cos\theta - \frac{1}{2}(-\cos\theta) = \frac{n+2}{2}\cos\theta

2. Simplification of the Denominator:

Let the denominator be DD: D=∑j=1n+1sin⁡2(jθ)=∑j=1n+11−cos⁡(2jθ)2=n+12−12∑j=1n+1cos⁡(2jθ)D = \sum_{j=1}^{n+1} \sin^2(j\theta) = \sum_{j=1}^{n+1} \frac{1 - \cos(2j\theta)}{2} = \frac{n+1}{2} - \frac{1}{2}\sum_{j=1}^{n+1} \cos(2j\theta)

Using the cosine series sum formula: ∑j=1n+1cos⁡(2jθ)=sin⁡((n+1)θ)sin⁡θcos⁡((n+2)θ)=sin⁡θsin⁡θcos⁡(π)=−1\sum_{j=1}^{n+1} \cos(2j\theta) = \frac{\sin((n+1)\theta)}{\sin\theta} \cos((n+2)\theta) = \frac{\sin\theta}{\sin\theta} \cos(\pi) = -1

Hence, the denominator becomes: D=n+12−12(−1)=n+22D = \frac{n+1}{2} - \frac{1}{2}(-1) = \frac{n+2}{2}

3. Expression for f(n)f(n):

f(n)=ND=n+22cos⁡θn+22=cos⁡θ=cos⁡(πn+2)f(n) = \frac{N}{D} = \frac{\frac{n+2}{2}\cos\theta}{\frac{n+2}{2}} = \cos\theta = \cos\left(\frac{\pi}{n+2}\right)

4. Evaluating the Options:

  • Option (A): f(4)=cos⁡(π4+2)=cos⁡(π6)=32f(4) = \cos\left(\frac{\pi}{4+2}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} Thus, Option (A) is correct.

  • Option (B): lim⁡n→∞f(n)=lim⁡n→∞cos⁡(πn+2)=cos⁡(0)=1≠12\lim_{n \to \infty} f(n) = \lim_{n \to \infty} \cos\left(\frac{\pi}{n+2}\right) = \cos(0) = 1 \neq \frac{1}{2} Thus, Option (B) is incorrect.

  • Option (C): f(6)=cos⁡(π8)f(6) = \cos\left(\frac{\pi}{8}\right) Since π8∈[0,π]\frac{\pi}{8} \in [0, \pi], we have cos⁡−1(f(6))=π8\cos^{-1}(f(6)) = \frac{\pi}{8}. α=tan⁡(π8)=2−1  ⟹  α+1=2\alpha = \tan\left(\frac{\pi}{8}\right) = \sqrt{2} - 1 \implies \alpha + 1 = \sqrt{2} Squaring both sides gives: (α+1)2=2  ⟹  α2+2α+1=2  ⟹  α2+2α−1=0(\alpha + 1)^2 = 2 \implies \alpha^2 + 2\alpha + 1 = 2 \implies \alpha^2 + 2\alpha - 1 = 0 Thus, Option (C) is correct.

  • Option (D): f(5)=cos⁡(π7)  ⟹  cos⁡−1(f(5))=π7f(5) = \cos\left(\frac{\pi}{7}\right) \implies \cos^{-1}(f(5)) = \frac{\pi}{7} sin⁡(7cos⁡−1f(5))=sin⁡(7⋅π7)=sin⁡(π)=0\sin\left(7 \cos^{-1} f(5)\right) = \sin\left(7 \cdot \frac{\pi}{7}\right) = \sin(\pi) = 0 Thus, Option (D) is correct.

Conclusion:

The correct options are A, C, and D.

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