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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll recall the half-angle identity for cosine function:

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

[cos (x/2)]^2 = [(1 + cos x)/2]

Let (a+b) = x such as:

[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}.

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