prove that cos 52 + cos 68 + cos172 = 0

Answer 1

first of all cos 172 = -cos 8

So now we have
LS
= cos 52 + cos 68 - cos 8

one of the lesser known identities is

cos A − cos B = −2 sin ½ (A + B) sin ½ (A − B)

(here is good list of them)
http://www.themathpage.com/atrig/trigonometric-identities.htm

let's look at our cos 52 - cos8, here let's A = 52, B=8
cos 52 - cos 8 = -2sin[60/2]sin[44/2]
cos52 -cos8 = - sin 22

so
cos 52 + cos 68 + cos172
= cos52 + cos68 - cos8
= cos52 - cos8 + cos68
= - sin22 + cos68

but by the complementary relationship
sin 22 = cos68 , (since 22+68 = 90)

so finally
= -cos 68 + cos 68
= 0
= RS

Q.E.D.

Answer 2

Prove that cos52°+cos68°+cos172=0