Asked by Anonymous
prove that cos 52 + cos 68 + cos172 = 0
Answers
Answered by
Reiny
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.
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.
Answered by
Pankaj
Prove that cos52°+cos68°+cos172=0
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