Question

If $ \cos ec\dfrac{\pi }{{32}} + \cos ec\dfrac{\pi }{{16}} + \cos ec\dfrac{\pi }{8} + \cos ec\dfrac{\pi }{4} + \cos ec\dfrac{\pi }{2} = \cot \dfrac{\pi }{k} $
Find k.

Answer 1

Hint: For this type of problem we will simplify left hand side of given equation by using term $ \cot \dfrac{\pi }{2} $ and grouping it with that cosec term which having same angle and then simplifying it by using trigonometric identity to get result and then using result obtain again in same manner till will completely simplify left hand side or convert it in term of cot then on comparing both side we will get value of k and hence required solution of given problem.
Formulas used: $ \cos ecx + \cot x = \cot \left( {\dfrac{x}{2}} \right) $

Complete step-by-step answer:
Given trigonometric equation is $ \cos ec\dfrac{\pi }{{32}} + \cos ec\dfrac{\pi }{{16}} + \cos ec\dfrac{\pi }{8} + \cos ec\dfrac{\pi }{4} + \cos ec\dfrac{\pi }{2} = \cot \dfrac{\pi }{k} $ …………….(i)
To find the value of ‘k’. We will simplify the left hand side of the given trigonometric equation.
Considering the left hand side of the given equation. We have,
 $ \cos ec\dfrac{\pi }{{32}} + \cos ec\dfrac{\pi }{{16}} + \cos ec\dfrac{\pi }{8} + \cos ec\dfrac{\pi }{4} + \cos ec\dfrac{\pi }{2} $
To simplify the above equation we add the value of $ \cot \dfrac{\pi }{2} $ in the above equation. It will not make any difference in the value of the equation as the value of $ \cot \dfrac{\pi }{2} $ is zero.
Therefore, we have
 $ \Rightarrow \cos ec\dfrac{\pi }{{32}} + \cos ec\dfrac{\pi }{{16}} + \cos ec\dfrac{\pi }{8} + \cos ec\dfrac{\pi }{4} + \cos ec\dfrac{\pi }{2} + \cot \dfrac{\pi }{2} $
Now, applying trigonometric identity $ \cos ecx + \cot x = \cot \left( {\dfrac{x}{2}} \right) $ on the last two terms of the above equation. WE have
 $ \Rightarrow \cos ec\dfrac{\pi }{{32}} + \cos ec\dfrac{\pi }{{16}} + \cos ec\dfrac{\pi }{8} + \cos ec\dfrac{\pi }{4} + \cot \dfrac{\pi }{4} $
Now, again applying the same identity on the last two terms of the above equation. We have,
 $ \cos ec\dfrac{\pi }{{32}} + \cos ec\dfrac{\pi }{{16}} + \cos ec\dfrac{\pi }{8} + \cot \dfrac{\pi }{8} $
Again applying the same identity on the last two terms. We have,
 $ \Rightarrow \cos ec\dfrac{\pi }{{32}} + \cos ec\dfrac{\pi }{{16}} + \cot \dfrac{\pi }{{16}} $
Again doing the same. We have
 $ \Rightarrow \cos ec\dfrac{\pi }{{32}} + \cot \dfrac{\pi }{{32}} $
Also, one more time doing the same.
 $ \Rightarrow \cot \dfrac{\pi }{{64}} $
Therefore, from above we see that value of left hand side of given trigonometric equation is $ \cot \dfrac{\pi }{{64}} $
 $ \Rightarrow \cos ec\dfrac{\pi }{{32}} + \cos ec\dfrac{\pi }{{16}} + \cos ec\dfrac{\pi }{8} + \cos ec\dfrac{\pi }{4} + \cos ec\dfrac{\pi }{2} = \cot \dfrac{\pi }{{64}}.....(ii) $
Then form above equation (i) and (ii). We have
 $ \Rightarrow \cot \dfrac{\pi }{k} = \cot \dfrac{\pi }{{64}} $
On comparing above trigonometric terms. We have,
 $ \Rightarrow k = 64 $
Hence, the required value of k is $ 64 $ .
So, the correct answer is “64”.

Note: For this type of problem if we apply any trigonometric identity on left hand side then it would not end in term of $ \cot \theta $ but there is a one half angle property with the help of which we can simplify or can find solution of given problem.
In this add $ \cot \dfrac{\pi }{2} $ in the left hand side so that we can use the identity $ \cos ecx + \cot x = \cot \left( {\dfrac{x}{2}} \right) $ . With the help of this identity we can simplify the left hand side with repeating the same steps again and again till we left with a single cotangent term which on equating with the right hand side gives out the value of k and hence the solution of given problem.