Prove that " cos^2 48 – sin^2 12 = (√5 + 1)/8 "

no use of trigonometric tables and calculators of any sort....... this is something related to sub multiple angles..... please anyone solve this for me....... please please
PLEASE DO THIS USING SUB MULTIPLE ANGLES PROPERTIES.........................

cos 2A= 2cos^2 A-1=1-2sin^2 A
so cos^2 48 – sin^2 12
=(1/2)(1+cos 96) - (1/2)(1-cos24)
=(1/2)(cos 96 +cos 24) and using cos(A)+cosB)=2cos((A+B)/2)cos((A-B)/2)
=cos60cos36
=(1/2)cos36
and you have to find cos36.
Suppose 5x=180, then
cos(5x)=cos(180) then x=-180,-108, -36, 36, 108
Also 3x=180-2x
so cos(3x)=cos(180-2x)=-cos2x
so 4cos^3(x) -3cos(x)=1 - 2cos^2(x)
giving 4c^3+2c^2 - 3c-1=0 where c=cosx
c=-1 is a root and factorizing gives
(c+1)(4c^2-2c-1)=0
so 4c^-2c-1=0 giving
c=(2±√20)/8=(1±√5)/4 and these have values cos(36) and cos(108)
the positive root is therefore cos(36)=(1+√5)/4
and the required value (1/2)cos(36)=(1+√5)/8

Very long method of solution. Quick Method:
cos²A - sin²B = cos(A + B)*cos(A - B)
So, cos²48 - sin²12 = cos(60)*cos(36) = (1/2)*(√5 + 1)/4 = (√5 + 1)/8

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