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Question: Prove that: \[2\cos\dfrac{\pi }{{13}} \cos\dfrac{{9\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}} + \cos\...

Prove that: 2cosπ13cos9π13+cos3π13+cos5π13=02\cos\dfrac{\pi }{{13}} \cos\dfrac{{9\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}} + \cos\dfrac{{5\pi }}{{13}} = 0

Explanation

Solution

We are given a trigonometric function, where we have to prove the term is equal to zero. We have, 2cosAcosB=cos(A+B)+cos(AB)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, 2cosπ13cos9π13+cos3π13+cos5π13=02\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.
2cosπ13cos9π13+cos3π13+cos5π132 \cos\dfrac{\pi }{{13}} \cos \dfrac{{9\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}
Using, 2cosAcosB=cos(A+B)+cos(AB)2\cos A \cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right) , we get,
cos(π13+9π13)+cos(π139π13)+cos3π13+cos5π13\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(10π13)+cos(8π13)+cos3π13+cos5π13= \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(θ)=cosθ\cos \left( { - \theta } \right) = \cos \theta , we get,
=cos(10π13)+cos(8π13)+cos3π13+cos5π13= \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(π3π13)+cos(π5π13)+cos3π13+cos5π13= \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}}
cos(πx)=cos(x)\because \cos (\pi - x) = - \cos (x) , we get,
=cos(3π13)cos(5π13)+cos3π13+cos5π13= - \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=0
R.H.S
2cosπ13cos9π13+cos3π13+cos5π13=02\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: [1,1]\left[ { - 1,1} \right]
  3. Period =2π= 2\pi
  4. x intercepts: x=π2+kπx = \dfrac{\pi }{2} + k\pi , where k is an integer.
  5. y intercepts: y=1y = 1
  6. maximum points: (2kπ,1)\left( {2k\pi ,1} \right) , where k is an integer.
  7. minimum points: (π+2kπ,1)\left( {\pi + 2k\pi , - 1} \right) , where k is an integer.
  8. symmetry: since cos(x)=cos(x)\cos \left( { - x} \right) = \cos \left( x \right) then cos(x)cos\left( x \right) is an even function and its graph is symmetric with respect to the y axis.