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Mathematics Question on Limits

For non-negative integers n, let f(n)=∑k=1nsin(k+1n+2π)sin(k+2n+2π)∑k=1nsin2(k+1n+2π)f\left(n\right) = \frac{\displaystyle\sum^{n}_{k = 1}sin\left(\frac{k+1}{n+2}\pi\right) sin \left(\frac{k+2}{n+2}\pi\right)}{\displaystyle\sum^{n}_{k = 1} sin^{2}\left(\frac{k+1}{n+2}\pi\right)} Assuming cos−1xcos^{-1}x takes values in [0,π]\left[0, \pi\right], which of the following options is/are correct?

A

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

B

limn→∞f(n)=12lim_{n\to\infty} f\left(n\right) = \frac{1}{2}

C

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

D

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

Answer

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

Explanation

Solution

f(n)=∑k=1n(cos(πn+2)−cos(2k+3n+2)π)∑k=1n(1−cos(2k+2n+2)π)f \left(n\right)=\frac{\displaystyle \sum_{k=1}^n\left(cos\left(\frac{\pi}{n+2}\right)-cos\left(\frac{2k+3}{n+2}\right)\pi\right)}{\displaystyle \sum_{k=1}^n\left(1-cos\left(\frac{2k+2}{n+2}\right)\pi\right)}
f(n)=(n+1)cos(πn+2)−(∑k=1ncos(2k+3n+2)π)(n+1)−(∑k=1ncos(2k+2n+2)π)f \left(n\right)=\frac{\left(n+1\right)cos\left(\frac{\pi}{n+2}\right)-\left(\displaystyle \sum_{k=1}^ncos\left(\frac{2k+3}{n+2}\right)\pi\right)}{\left(n+1\right)-\left(\displaystyle \sum_{k=1}^ncos\left(\frac{2k+2}{n+2}\right)\pi\right)}
f(n)=(n+1)cosπn+2−(sin((n+1)πn+2)sin(πn+2).cos(n+3n+2)π)(n+1)−(sin((n+1)ππn+2)sin(πn+2).cos(2(n+2)π2(n+2)))f \left(n\right)=\frac{\left(n+1\right)cos \frac{\pi}{n+2}-\left(\frac{sin \left(\frac{\left(n+1\right)\pi}{n+2}\right)}{sin \left(\frac{\pi}{n+2}\right)}.cos \left(\frac{n+3}{n+2}\right)\pi\right)}{\left(n+1\right)-\left(\frac{sin\left(\frac{\left(n+1\right)\pi\pi}{n+2}\right)}{sin \left(\frac{\pi}{n+2}\right)}.cos \left(\frac{2\left(n+2\right)\pi}{2\left(n+2\right)}\right)\right)}
f(n)=(n+1)cos(πn+2)+cos(πn+2)(n+1)+1⇒g(x)=cos(πn+2)f \left(n\right)=\frac{\left(n+1\right)cos\left(\frac{\pi}{n+2}\right)+cos\left(\frac{\pi}{n+2}\right)}{\left(n+1\right)+1} \Rightarrow g\left(x\right)=cos\left(\frac{\pi}{n+2}\right)
(A)sin(7 cos−1 π7)=sin π=0\left(A\right) sin\left(7\,cos^{-1}\, \frac{\pi}{7}\right)=sin\,\pi=0
(B)f(4)=cosπ6=32\left(B\right) f \left(4\right)=cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}
(C)\left(C\right) lim⁡n→∞\displaystyle \lim_{n \to \infty} cos(πn+2)=1cos\left(\frac{\pi}{n+2}\right)=1
(D)α=tan(cos−1 cosπ8)=2−1⇒α+12\left(D\right) \alpha=tan\left(cos^{-1}\,cos \frac{\pi}{8}\right)=\sqrt{2}-1 \Rightarrow \alpha+1 \sqrt{2}
α2+2α−1=0\alpha^{2}+2\alpha-1=0