Question

Prove that: \[2\cos\dfrac{\pi }{{13}} \cos\dfrac{{9\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}} + \cos\dfrac{{5\pi }}{{13}} = 0\]

Answer 1

Hint:
We are given a trigonometric function, where we have to prove the term is equal to zero. We have, \[2 \cos A \cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)\], which we are going to use in the given problem to simplify the problem. Then we use normal properties of cosine function to get to our result.

Complete step by step solution:
We are here given, \[2\cos \dfrac{\pi }{{13}} \cos \dfrac{{9\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}} = 0\]
So, we here start with,
L.H.S.
 \[2 \cos\dfrac{\pi }{{13}} \cos \dfrac{{9\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}\]
Using, \[2\cos A \cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)\] , we get,
 \[ \Rightarrow \cos \left( {\dfrac{\pi }{{13}} + \dfrac{{9\pi }}{{13}}} \right) + \cos \left( {\dfrac{\pi }{{13}} - \dfrac{{9\pi }}{{13}}} \right) + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}\]
Now, on simplifying, we get,
 \[ = \cos \left( {\dfrac{{10\pi }}{{13}}} \right) + \cos \left( { - \dfrac{{8\pi }}{{13}}} \right) + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}\] ​
Using, \[\cos \left( { - \theta } \right) = \cos \theta \], we get,
 \[ = \cos \left( {\dfrac{{10\pi }}{{13}}} \right) + \cos \left( {\dfrac{{8\pi }}{{13}}} \right) + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}\]
Now, again, simplifying, we get,
 \[ = \cos \left( {\pi - \dfrac{{3\pi }}{{13}}} \right) + \cos \left( {\pi - \dfrac{{5\pi }}{{13}}} \right) + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}\]
 \[\because \cos (\pi - x) = - \cos (x)\] , we get,
 \[ = - \cos \left( {\dfrac{{3\pi }}{{13}}} \right) - \cos \left( {\dfrac{{5\pi }}{{13}}} \right) + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}\]
Cancelling out, we get,
 $=0$
R.H.S
 \[2\cos \dfrac{\pi }{{13}}\cos \dfrac{{9\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}} = 0\]
Hence, proved.

Note:
The properties of the cosine function are,
1) Domain: all real numbers
2) Range: \[\left[ { - 1,1} \right]\]
3) Period \[ = 2\pi \]
4) x intercepts: \[x = \dfrac{\pi }{2} + k\pi \] , where k is an integer.
5) y intercepts: \[y = 1\]
6) maximum points: \[\left( {2k\pi ,1} \right)\] , where k is an integer.
7) minimum points: \[\left( {\pi + 2k\pi , - 1} \right)\] , where k is an integer.
8) symmetry: since \[\cos \left( { - x} \right) = \cos \left( x \right)\] then \[cos\left( x \right)\] is an even function and its graph is symmetric with respect to the y axis.