Question

If $ f(n) = 2\cos nx\,\forall \,n\, \in \,N,\,then\,f(1)f(n + 1) - f(n) $ is equal to
(A) $ f(n + 3) $
(B) $ f(n + 2) $
(C) $ f(n + 1)f(2) $
(D) $ f(n + 2)(f(2) $

Answer 1

Hint: First we used a given function f(n) to find values of different functions like f(1), f(n+1) and f(n) and then using these functions together proceeded with some trigonometric formulas to do simplification and get required correct option.
 $ 2\cos A.\cos B = \cos (A + B) + \cos (A - B),\,\, $ and cosine is an even function therefore $ \cos ( - \theta ) = \cos \theta $

Complete step-by-step answer:
Here, given function in problems is
 $ f(n) = 2\cos nx $
Then to find value of $ f(1)f(n + 1) - f(n) $ we calculate, first values of f (1), f (n+1) and f(n) and then using these values to
Check which one of the given options is correct.
 $ \therefore $ f (1) = $ 2\cos (1.x) $
 $ \Rightarrow f(1) = 2\cos x $ …… (i)
Now, calculating f (n+1) we have
 $ f(n + 1) = 2\cos (n + 1)x $ ……. (ii)
And $ f(n) = 2\cos nx $ ……. (iii)
Using from (i), (ii) and (iii) we have
 $ f(1)f(n + 1) - f(n) = 2\cos x.2\cos (n + 1)x - 2\cos nx $
 $
   = 4.\cos x.\cos (n + 1)x - 2\cos nx \\
   = 2\left\{ {2\cos x.\cos (n + 1)x} \right\} - 2\cos nx \\
   = 2\left\{ {\cos (x + nx + x) + \cos (x - nx - x} \right\} - 2\cos nx \\
   = 2\left\{ {\cos (n + 2)x + \cos ( - nx)} \right\} - 2\cos nx \\
   = 2\{ \cos (n + 2)x + \cos (nx)\} - 2\cos nx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\,\because \cos ( - nx) = \cos nx} \right) \\
   = 2\cos (n + 2)x + 2\cos nx - 2\cos nx \\
   = 2\cos (n + 2)x \\
  $
 $ \Rightarrow f(1)f(n + 1) - f(n) = 2\cos (n + 2)x $

Therefore, from above we see that the value of $ f(1)f(n + 1) - f(n) $ is $ 2\cos (n + 2)x $ . Which can be written as $ f(n + 2) $ .
Hence, from given four options we see that option (B) is the correct option.

So, the correct answer is “Option B”.

Note: As, we know that every function given values according to domains we are taking. So, changing the domain for the same functions given different results. Hence, to get different values for the same function we just replace the given function with required domains to get required results.