Question

Question #34291

Answer 1

`sin^4x+cos^4x`

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

`=1-2sin^2xcos^2x`

`=1-1/2xx4sin^2xcos^2x`

`=1-1/2xx(2sinxcosx)^2`

`=1-1/2xx((sinx+cosx)^2-(sin^2x+cos^2))^2`

`=1-1/2xx(a^2-1)^2`

`=(2-(a^2-1)^2)/2`

`=(2-a^4+2a^2-1)/2`

`=(1-a^4+2a^2)/2`