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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
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(1+2cos(90-18))
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
Therefore LHS=RHS
Hence Proved!
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