show that cos12+cos60+cos84=cos24+cos48

=cos12+2cos72cos12

=cos12(1+2cos72)

=cos12(1+2sin18)

=cos12(1+2(((5^(1/2))-1)/4)

=cos12*2*(5^0.5+1)/4

=2cos12cos36

=cos48+cos24

Hi, Ranjith

L.H.S.

=cos 12+cos 60+cos 84

=cos 12+(cos 84+cos 60)

=cos 12+2.cos 72 . cos 12

=(1+2sin 18)cos 12

=(1+2.(√5 -1)/4)cos 12

=(1+.(√5 -1)/2)cos 12

=(√5 +1)/2.cos 12

R.H.S

=cos 48+cos 24

=2cos 36. cos 12

=2.(√5 + 1)/4)cos 12

=(√5 +1)/2.cos 12

L.H.S = R.H.S

So,cos12+cos60+cos84=cos24+cos48

LHS

cos12+(cos60+cos84)= cos12+2cos(60+84/2)cos(60-84/2)

=cos12+2cos72cos12

=cos12(1+2cos72)

=cos12(1+2cos(90-18))

=cos12(1+2sin18)

put the value of sin18 in the above equation. (We know that sin18= root5-1/4)

=(root5+1 whole divided by 2 )* cos12

RHS

cos24+cos48=2cos36cos12

Again by substituting the value of cos36=root5+1/4

we get

=(root5+1 whole divided by 2 )* cos12

Therefore LHS=RHS

Hence Proved!