For non-negative integers n, let fn=∑k=0nsin⁡k+1n+2πsin⁡k+2n+2π∑k=0nsin2⁡k+1n+2π Assuming cos-1⁡x takes values in 0, π, which of the following options is/are correct?

(D) If α=tan⁡cos-1⁡f6, then α2+2α-1=0

JEE Advanced · Mathematics · 7. Trigonometry

For non-negative integers $n,$ let
$f (n) = \frac{\sum_{k = 0}^{n} \sin (\frac{k + 1}{n + 2} π) \sin (\frac{k + 2}{n + 2} π)}{\sum_{k = 0}^{n} \sin^{2} (\frac{k + 1}{n + 2} π)}$
Assuming $\cos^{- 1} x$ takes values in $[0, π],$ which of the following options is/are correct?

A $\sin (7 \cos^{- 1} f (5)) = 0$
B $f (4) = \frac{\sqrt{3}}{2}$
C $\lim_{n \to \infty} f (n) = \frac{1}{2}$
D If $α = \tan (\cos^{- 1} f (6)),$ then $α^{2} + 2 α - 1 = 0$

Verified Solution

Answer & Solution

Correct Answer

(A) $\sin (7 \cos^{- 1} f (5)) = 0$

Step-by-step Solution

Detailed explanation

$f(n)=\frac{\sum_{k=0}^n \sin \left(\left(\frac{k+1}{n+2}\right) \pi\right) \sin \left(\left(\frac{k+2}{n+2}\right) \pi\right)}{\sum_{k=0}^n \sin ^2\left(\left(\frac{k+1}{n+2}\right) \pi\right)}$
$(\because 2 \sin C \sin D=\cos (C-D)$ $-~\cos (C+D))$
$\because 2 \sin ^2 \theta=1-\cos 2 \theta$

f (n) = \frac{\sum_{k = 0}^{n} (\cos (\frac{π}{n + 2}) - \cos (\frac{2 k + 3}{n + 2} π))}{\sum_{k = 0}^{n} (1 - \cos ((\frac{2 k + 2}{n + 2}) π))}

f (n) = \frac{(n + 1) \cos (\frac{π}{n + 2}) - \sum_{k = 0}^{n} \cos (\frac{2 k + 3}{n + 2}) π}{(n + 1) - \sum_{k = 0}^{n} \cos (\frac{2 k + 2}{n + 2}) π}

Now $\cos \alpha+\cos (\alpha+\beta)+\ldots+\cos (\alpha~+$ $(n-1) \beta)=\frac{\sin \left(\frac{n \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} \cos \left(\alpha+\frac{(n-1) \beta}{2}\right)$

f (n) = \frac{(n + 1) \cos (\frac{π}{n + 2}) - \frac{\sin (\frac{n + 1}{n + 2} π)}{\sin (\frac{π}{n + 2})} \cos (\frac{n + 3}{n + 2} π)}{(n + 1) - \frac{\sin (\frac{n + 1}{n + 2}) π}{\sin (\frac{π}{n + 2})} \cos (π)}

In numerator cosine sum

α = \frac{3 π}{n + 2}

β = \frac{2 π}{n + 2}

Number of terms

= (n + 1)

In denominator cosine sum

α = \frac{2 π}{n + 2}

β = \frac{2 π}{n + 2}

Number of terms

= (n + 1)

Now

\sin (\frac{n + 1}{n + 2} π) = \sin (π - \frac{π}{n + 2}) = \sin \frac{π}{n + 2}

\cos (\frac{n + 3}{n + 2} π) = \cos (π + \frac{π}{n + 2}) = - \cos \frac{π}{n + 2}

f (n) = \frac{(n + 1) \cos (\frac{π}{n + 2}) + \cos (\frac{π}{n + 2})}{(n + 1) - (- 1)}

f (n) = \frac{\cos (\frac{π}{n + 2}) (n + 2)}{(n + 2)}

f (n) = \cos (\frac{π}{n + 2})

Hence,

(A) f (4) = \cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}

(B) \lim_{n + \infty} f (n) = \cos (0) = 1

(C) f (6) = \cos (\frac{π}{8}) :

then

α = \tan (\frac{π}{8}) = \sqrt{2} - 1

α = \sqrt{2} (5) = \cos (\frac{π}{7})

\sin (7 \cos^{- 1} (\cos \frac{π}{7})) = \sin π = 0

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From JEE Advanced

Explore more questions on app

For non-negative integers n, letfn=∑k=0nsin⁡k+1n+2πsin⁡k+2n+2π∑k=0nsin2⁡k+1n+2πAssuming cos-1⁡x takes values in 0, π, which of the following options is/are correct?

Step-by-step Solution

For non-negative integers $n,$ let
$f (n) = \frac{\sum_{k = 0}^{n} \sin (\frac{k + 1}{n + 2} π) \sin (\frac{k + 2}{n + 2} π)}{\sum_{k = 0}^{n} \sin^{2} (\frac{k + 1}{n + 2} π)}$
Assuming $\cos^{- 1} x$ takes values in $[0, π],$ which of the following options is/are correct?