Let as above $x = cos θ + isinθ , then 1/x = cos θ- isinθ$

$(2cosθ )^6= (x + 1/x )^6$

$ = x^6 + 6x^5.1/x + 15x^4.1/x^2 + 20x^3.1/x^3 + 15x^2.1/x^4 + 6x^1.1/x^5 +1/x^6$

$ = x^6 + 6x^5 + 15x^2 + 20 + 151/x^2 + 61/x^4 +1/x^6 …………………….(1)$

$(2isinθ )^6= (x–1/x )^6$

$ = x^6– 6x^5 + 15x^2 - 20 + 151/x^2 – 61/x^4 +1/x^6 ……………………..(2)$ $(2sinθ )^6= x^6+ 6x^5– 15x^2 + 20– 151/x^2 + 61/x^4 - 1/x^6 $

Subtracting (2) from (1),

$2^6 (cos)^6 θ-(sin)^6 θ)= [x^6 + 6x^5 + 15x^2 + 20 + 151/x^2 + 61/x^4 +1/x^6 ] – $

$[- x^6 + 6x^5– 15x^2 + 20– 151/x^2 + 61/x^4 - 1/x^6 ]$

$ = 2(x^6+ 1/x^6 ) + 15(x^2+ 1/x^2 )$

$ = 2cos6θ + 15cos2θ ………………….[ (x^6+ 1/x^6 ) = cos6θ]$

$∴2^6 ((cos)^6 θ-(sin)^6 θ) = 2cos6θ + 15cos2θ.$

$(cos)^6 θ-(sin)^6 θ= 1/16 (cos6θ+15cos2θ)$