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

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

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.

Approved by eNotes Editorial Team