Question

sin^248-cos^212=√5+1/8

Answer 1

I think u mean
cos^2 48 – sin^2 12 = (√5 + 1)/8  
your answer is
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