If `sin x+cos x=a ,` find the value of `sin^6x+cos^6xdot`

Explanation

`sin^6 x+cos^6 x=(sin^2 x)^3+(cos^3 x)^3`

`=(sin^2 x+cos^2 x)^3-3sin^2 x cos^2 x(sin^2 x+cos^2 x)`

`=1^3-3sin^2x cos^2x`

Using the formula `sin^2x+cos^2x=1`

`=1-3{frac{(sin x+cos x)^2-sin^2x-cos^2 x}{2}}^2`

`=1-3{frac{a^2-(sin^2x+cos^2x)}{2}}^2`

`=1-3{frac{a^2-1}{2}}^2`

`=frac{1}{4}{4-3(a^2-1)^2}`