Question
Question: Prove that: \[2\cos\dfrac{\pi }{{13}} \cos\dfrac{{9\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}} + \cos\...
Prove that: 2cos13πcos139π+cos133π+cos135π=0
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(A−B), 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, 2cos13πcos139π+cos133π+cos135π=0
So, we here start with,
L.H.S.
2cos13πcos139π+cos133π+cos135π
Using, 2cosAcosB=cos(A+B)+cos(A−B) , we get,
⇒cos(13π+139π)+cos(13π−139π)+cos133π+cos135π
Now, on simplifying, we get,
=cos(1310π)+cos(−138π)+cos133π+cos135π
Using, cos(−θ)=cosθ, we get,
=cos(1310π)+cos(138π)+cos133π+cos135π
Now, again, simplifying, we get,
=cos(π−133π)+cos(π−135π)+cos133π+cos135π
∵cos(π−x)=−cos(x) , we get,
=−cos(133π)−cos(135π)+cos133π+cos135π
Cancelling out, we get,
=0
R.H.S
2cos13πcos139π+cos133π+cos135π=0
Hence, proved.
Note:
The properties of the cosine function are,
- Domain: all real numbers
- Range: [−1,1]
- Period =2π
- x intercepts: x=2π+kπ , where k is an integer.
- y intercepts: y=1
- maximum points: (2kπ,1) , where k is an integer.
- minimum points: (π+2kπ,−1) , where k is an integer.
- symmetry: since cos(−x)=cos(x) then cos(x) is an even function and its graph is symmetric with respect to the y axis.