Verify the identity cos^2(a+b) + cos^2(a-b)= 1 + cos2a*cos2b

We'll write the formula for cosine of a
half-angle:

cos (x/2) = sqrt[(1 + cos
x)/2]

[cos (x/2)]^2 = [(1 + cos
x)/2]

We'll put (a+b) = x
so:

[cos(a+b)]^2 = [(1 + cos 2(a+b))/2]
(1)

[cos(a-b)]^2 = [(1 + cos 2(a-b))/2]
(2)

We'll add (1) + (2):

[(1 +
cos 2(a+b) + 1 + cos 2(a-b)/2] = 1 + cos 2a*cos 2b

2/2 +
[cos 2(a+b)]/2 +  [cos 2(a-b)]/2 = 1 + cos 2a*cos 2b

We'll
eliminate 1 both sides:

[cos 2(a+b)]/2 +  [cos 2(a-b)]/2 =
cos 2a*cos 2b

cos 2(a+b) +  cos 2(a-b) = 2cos 2a*cos
2b

cos 2(a+b) = 2[cos (a+b)]^2 -
1

cos 2(a-b) = 2[cos (a-b)]^2 -
1

cos 2a = 2 (cos a)^2 - 1

cos
2b = 2 (cos b)^2 - 1

2cos 2a*cos 2b = 2[2 (cos a)^2 - 1]*[2
(cos b)^2 - 1]

2cos 2a*cos 2b = 2{4(cos a)^2*(cos
b)^2-2[(cos a)^2+(cos b)^2] + 1}

[cos (a+b)]^2 = [cos
(a+b)][cos (a+b)] =
(cosa*cosb-sina*sinb)^2

(cosa*cosb-sina*sinb)^2 =
(cosa*cosb)^2 - 2cosa*cosb*sina*sinb +
(sina*sinb)^2

2(cosa*cosb-sina*sinb)^2 = 2(cosa*cosb)^2 -
4cosa*cosb*sina*sinb + 2(sina*sinb)^2

2[cos (a+b)]^2 - 1 =
2(cosa*cosb)^2 - 4cosa*cosb*sina*sinb + 2(sina*sinb)^2 - 1
(3)

2[cos (a-b)]^2 - 1 =2(cosa*cosb)^2 +
4cosa*cosb*sina*sinb + 2(sina*sinb)^2 - 1 (4)

We'll add (3)
+ (4):

2(cosa*cosb)^2 - 4cosa*cosb*sina*sinb +
2(sina*sinb)^2 - 1 +2(cosa*cosb)^2 + 4cosa*cosb*sina*sinb + 2(sina*sinb)^2 -
1

We'll eliminate like
terms:

 4(cosa*cosb)^2 + 4(sina*sinb)^2 - 2 = 2{4(cos
a)^2*(cos b)^2-2[(cos a)^2+(cos b)^2] + 1}