Using De Moivre’s theorem prove that]

`cos^6theta-sin^6theta=1/16(cos6theta+15cos2theta)`

#### Solution

Let as above x=cosθ+isinθ, then `1/x=costheta-isintheta`

`(2costheta)^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+15 1/x^2+6 1/x^4+ 1/x^6`…………………….(1)

`(2isintheta)^6=(x-1/x)^6`

`=x^6-6x^5+15x^2-20+15 1/x^2-6 1/x^4+1/x^6` ...................(2)

`(2sintheta)^6=x^6+6x^5-15x^2+20-15 1/x^2+6 1/x^4-1/x^6`

Subtracting (2) from (1),

`2^6(cos^6theta-sin^6theta)=[x^6+6x^5+15x^2+20+15 1/x^2+6 1/x^4+1/x^6]-[-x^6+6x^5-15x^2+20-15 1/x^2+6 1/x^4-1/x^6]`

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

`=2cos6theta+15cos2theta..................[(x^6+1/x^6)=cos6theta]`

`therefore 2^6(cos^6theta-sin^6theta)=2cos6theta+15cos2theta`

`cos^6theta-sin^6theta=1/16(cos6theta+15cos2theta)`