Expand cosh^8θ in terms of hyperbolic cosines of θ

Recall that
cosh(x) = [ e^(x) + e^(-x) ]/2
Then
cosh(2x) = [ e^(2x) + e^(-2x) ]/2
cosh(4x) = [ e^(4x) + e^(-4x) ]/2
cosh(6x) = [ e^(6x) + e^(-6x) ]/2
cosh(8x) = [ e^(8x) + e^(-8x) ]/2

By Binomial Expansion
(a + b)^8
= a^8 + 8a^7 b + 28a^6 b^2 + 56a^5 b^3 + 70a^4 b^4 + 56a^3 b^5 + 28a^2 b^6 + 8ab^7 + b^8
= (a^8 + b^8) + 8ab(a^6 + b^6) + 28a^2 b^2 (a^4 + b^4) + 56a^3 b^3 (a^2 + b^2) + 70a^4 b^4
= (a^8 + b^8) + 8ab(a^6 + b^6) + 28(ab)^2 (a^4 + b^4) + 56(ab)^3 (a^2 + b^2) + 70(ab)^4

Letting a = e^(x), b = e^(-x), we have
ab = 1
a + b = 2cosh(x)
a^8 + b^8 = 2cosh(8x)
a^6 + b^6 = 2cosh(6x)
a^4 + b^4 = 2cosh(4x)

Substitute them into the above equation gives
2^8 * cosh⁸(x)
= 2cosh(8x) + 8 * 2cosh(6x) + 28 * 2cosh(6x) + 56 * 2cosh(2x) + 70
= 2cosh(8x) + 16cosh(6x) + 56cosh(6x) + 112cosh(2x) + 70

==> 2^8 * cosh⁸(x) = 2cosh(8x) + 16cosh(6x) + 56cosh(6x) + 112cosh(2x) + 70
==> 2^7 * cosh⁸(x) = cosh(8x) + 8cosh(6x) + 28cosh(6x) + 56cosh(2x) + 35
==> 128 cosh⁸(x) = cosh(8x) + 8cosh(6x) + 28cosh(6x) + 56cosh(2x) + 35
==> cosh⁸(x) = (1/128) [ cosh(8x) + 8cosh(6x) + 28cosh(6x) + 56cosh(2x) + 35 ]

Typo sorry 'bout that.

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