Value of cos^2 (48) - sin^2 (12) ?

The value of that (√5 + 1) / 8 and We can prove it like this:

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

I think this can be very useful