Question

For non-negative integer n, let${\text{f}}\left( {\text{n}} \right) = \dfrac{{\sum\limits_{{\text{k - 0}}}^{\text{n}} {{\text{sin}}\left( {\dfrac{{{\text{k + 1}}}}{{{\text{n + }}2}}\pi } \right)} {\text{ sin}}\left( {\dfrac{{{\text{k + 2}}}}{{{\text{n + }}2}}\pi } \right)}}{{\sum\limits_{{\text{k - 0}}}^{\text{n}} {{\text{sin}}\left( {\dfrac{{{\text{k + 1}}}}{{{\text{n + }}2}}\pi } \right)} }}$. Assuming ${\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}$takes values in [0, π], which of the following options is/are correct?
This question has multiple correct options.
$
  {\text{A}}{\text{. f}}\left( 4 \right) = \dfrac{{\sqrt 3 }}{2} \\
  {\text{B}}{\text{. If }}\alpha {\text{ = tan}}\left( {{\text{co}}{{\text{s}}^{ - 1}}{\text{f}}\left( 6 \right)} \right){\text{ then }}{\alpha ^2} + 2\alpha - 1 = 0 \\
  {\text{C}}{\text{. sin}}\left( {7{\text{co}}{{\text{s}}^{ - 1}}{\text{f}}\left( 5 \right)} \right) = 0 \\
  {\text{D}}{\text{. }}{}_{{\text{n}} \to \infty }\left( {\text{n}} \right) = \dfrac{1}{2} \\
$

Answer 1

Hint:In order to find which of the options are correct we first simplify the given function f (n) and then verify each of the options individually to check if they are correct. We use the formulae of trigonometric functions such as “Cos A – Cos B” to simplify the given function and then substitute its limits to find the value of f (n).

Complete step by step answer:
Given Data,
${\text{f}}\left( {\text{n}} \right) = \dfrac{{\sum\limits_{{\text{k - 0}}}^{\text{n}} {{\text{sin}}\left( {\dfrac{{{\text{k + 1}}}}{{{\text{n + }}2}}\pi } \right)} {\text{ sin}}\left( {\dfrac{{{\text{k + 2}}}}{{{\text{n + }}2}}\pi } \right)}}{{\sum\limits_{{\text{k - 0}}}^{\text{n}} {{\text{sin}}\left( {\dfrac{{{\text{k + 1}}}}{{{\text{n + }}2}}\pi } \right)} }}$
${\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}$takes values from the set [0, π]
n is a non-negative integer.

We know the formula of a trigonometric term of the form Cos A – Cos B =$ - 2{\text{ sin}}\dfrac{{{\text{A + B}}}}{2}{\text{ sin}}\dfrac{{{\text{A - B}}}}{2}$.
Using this identity we start simplifying the given function f (n) and then we substitute its upper and lower limits, as follows:
\[
  {\text{f}}\left( {\text{n}} \right) = \dfrac{{\sum\limits_{{\text{k - 0}}}^{\text{n}} {{\text{sin}}\left( {\dfrac{{{\text{k + 1}}}}{{{\text{n + }}2}}\pi } \right)} {\text{ sin}}\left( {\dfrac{{{\text{k + 2}}}}{{{\text{n + }}2}}\pi } \right)}}{{\sum\limits_{{\text{k - 0}}}^{\text{n}} {{\text{sin}}\left( {\dfrac{{{\text{k + 1}}}}{{{\text{n + }}2}}\pi } \right)} }} \\
   \Rightarrow \dfrac{{\sum\limits_{{\text{k - 0}}}^{\text{n}} {\left( {{\text{cos}}\left( {\dfrac{\pi }{{{\text{n + }}2}}} \right) - {\text{sin}}\left( {\dfrac{{{\text{2k + 3}}}}{{{\text{n + }}2}}\pi } \right)} \right)} }}{{\sum\limits_{{\text{k - 0}}}^{\text{n}} {{\text{2si}}{{\text{n}}^2}\left( {\dfrac{{{\text{k + 1}}}}{{{\text{n + }}2}}\pi } \right)} }} \\
  {\text{Now we substitute the limits}} \\
   \Rightarrow \dfrac{{\left( {{\text{n + 1}}} \right){\text{cos}}\dfrac{\pi }{{{\text{n + }}2}} - \dfrac{{{\text{cos}}\left( {\dfrac{{{\text{n + 3}}}}{{{\text{n + }}2}}} \right)\pi {\text{ sin}}\left( {\dfrac{{{\text{n + 1}}}}{{{\text{n + }}2}}} \right)\pi }}{{{\text{sin}}\dfrac{\pi }{{{\text{n + }}2}}}}}}{{\left( {{\text{n + 1}}} \right) - \dfrac{{{\text{cos}}\pi {\text{ sin}}\left( {\dfrac{{{\text{n + 1}}}}{{{\text{n + }}2}}} \right)\pi }}{{{\text{sin}}\dfrac{\pi }{{{\text{n + }}2}}}}}} \\
   \Rightarrow \dfrac{{\left( {{\text{n + 1}}} \right){\text{cos}}\dfrac{\pi }{{{\text{n + }}2}} + {\text{cos}}\left( {\dfrac{{{\text{n + 3}}}}{{{\text{n + 3}}}}} \right)\pi }}{{\left( {{\text{n + 1}}} \right) + 1}} \\
   \Rightarrow \dfrac{{\left( {{\text{n + 1}}} \right){\text{cos}}\left( {\dfrac{\pi }{{{\text{n + }}2}}} \right) + {\text{cos}}\left( {\dfrac{\pi }{{{\text{n + }}2}}} \right)}}{{{\text{n + 2}}}} \\
   \Rightarrow {\text{cos}}\left( {\dfrac{\pi }{{{\text{n + }}2}}} \right) \\
\]
Hence we obtained the value of f (n): \[{\text{f}}\left( {\text{n}} \right){\text{ = cos}}\left( {\dfrac{\pi }{{{\text{n + }}2}}} \right)\]

Now we compute and verify each of the given options w.r.t,\[{\text{f}}\left( {\text{n}} \right){\text{ = cos}}\left( {\dfrac{\pi }{{{\text{n + }}2}}} \right)\]
${\text{f}}\left( 4 \right) = {\text{cos}}\left( {\dfrac{\pi }{{{\text{n + }}2}}} \right) = {\text{cos}}\left( {\dfrac{\pi }{{{\text{4 + }}2}}} \right) = {\text{cos}}\left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2}$
Option A is correct
$\alpha {\text{ = tan}}\left( {{\text{co}}{{\text{s}}^{ - 1}}{\text{f}}\left( 6 \right)} \right){\text{ = tan}}\left( {{\text{co}}{{\text{s}}^{ - 1}}\left( {{\text{cos}}\dfrac{\pi }{8}} \right)} \right){\text{ = tan}}\dfrac{\pi }{8}$
${\text{Now let us compute, tan}}\dfrac{\pi }{4} = \dfrac{{2{\text{tan}}\dfrac{\pi }{8}}}{{1 - {\text{ta}}{{\text{n}}^2}\dfrac{\pi }{8}}} \Rightarrow 1 = \dfrac{{2\alpha }}{{1 - {\alpha ^2}}} \Rightarrow {\alpha ^2} + 2\alpha - 1 = 0$
Option B is not correct
${\text{sin}}\left( {7{\text{co}}{{\text{s}}^{ - 1}}{\text{f}}\left( 5 \right)} \right) = {\text{sin}}\left( {7{\text{co}}{{\text{s}}^{ - 1}}\left( {{\text{cos}}\dfrac{\pi }{7}} \right)} \right){\text{ = sin}}\pi = 0$
Option C is correct
$\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } {\text{f}}\left( {\text{n}} \right) = {\text{cos}}\left( {\dfrac{\pi }{{{\text{n + }}2}}} \right) = {\text{cos}}\left( {\dfrac{\pi }{\infty }} \right) = {\text{cos 0 = 1}}$
Option D is correct.

Hence the options A, C, D are the correct answers.

Note: In order to solve this type of question the key is to know how to simplify the given function. We used trigonometric identities to solve the function and further simplify it by computing its limits. We have to be very careful while performing this as the calculation is lengthy and also tricky.
We obtain the trigonometric values of cos and tan functions by referring to their respective trigonometric tables.
Any fraction with a denominator of infinity is zero.