# Prove that: `2cos(pi/13)cos({9pi}/13)+cos({3pi}/13)+cos({5pi}/13)=0`

### 2 Answers | Add Yours

To prove that this equation is always true, we use the trigonometric product-to-sum identity (given in the website below):

`2cos({a-b}/2)cos({a+b}/2)=cosa+cosb`

Letting `a={5pi}/13` and `b={3pi}/13` , we get:

`LS=2cos(pi/13)cos({9pi}/13)+cos({3pi}/13)+cos({5pi}/13)`

`=2cos(pi/13)cos({9pi}/13)+2cos(pi/13)cos({4pi}/13)`

`=2cos(pi/13)(cos({9pi}/13)+cos({4pi}/13))` now use the identity again

`=2cos(pi/13)2cos({5pi}/26)cos(pi/2)` but `cos(pi/2)=0`

`=0=RS`

**The equation is proven true.**

**Sources:**

Require to prove :

2 cos(pi/13) cos(9pi/13) + cos(3pi/13)+cos(5 pi/13) = 0

Let us take L.H.S

L.H.S=> 2 cos(pi/13) cos(9 pi/13) + cos(3 pi/13)+cos(5 pi/13)

=> 2 cos(pi/13) cos(9 pi/13) + cos(3 pi/13)+cos(5 pi/13)

=>2cos(pi/13)cos(9 pi/13)+2cos((3 pi+5 pi)/2)/13)cos((5pi-3pi/13)

[ By applying : cosA+ cosB = 2cos((A+B)/2) cos((A-B)/2) ]

=>2 cos(pi/13) cos(9 pi/13) + 2 cos((8 pi/2)/13) cos((2 pi/2)/13)

=> 2 cos(pi/13) cos(9 pi/13) + 2cos(4 pi/13) cos(pi/13)

[Since, (3 pi +5 pi)/2=8 pi /2 = 4 pi & (5 pi – 3 pi)/2 = 2 pi/2= pi ]

=> 2 cos(pi/13) cos(9 pi/13) + 2 cos(pi- (9pi/13)) cos(pi/13)

[ Since, cos(4 pi/13) = cos(pi- 9 pi/13) ]

=> 2 cos(pi/13) cos(9 pi/13) - 2 cos(pi/13) cos(9 pi/13)

[ since cos(180-A) = - cosA]

=> 0 = R.H.S

L.H.S = R.H.S

**Hence, **

**2 cos(pi/13) cos(9pi/13) + cos(3pi/13)+cos(9pi/13) = 0 **** Proved**** **