If `sinx+cosx=a`, then
(i) `sin^(6)x+cos^(6)x=.....`
(ii) `abs(sinx-cosx)=.....`

Explanation

Given that, `sinx+cosx=a`
On squaring both sides, we get
`" "(sinx+cosx)^(2)=(a)^(2)`
`rArr" "sin^(2)x+cos^(2)x+2sinxcosx=a^(2)`
`rArr" "sinx*cosx=(1)/(2)(a^(2)-1)`
`(i) sin^(6)x+cos^(6)x=(sin^(2)x)^(3)+(cos^(2)x)^(3)`
`" "=(sin^(2)x+cos^(2)x)(sin^(4)x-sin^(2)xcos^(2)x+cos^(4)x)` ltBrgt `" "=sin^(4)x+cos^(4)x-(1)/(4)(a^(2)-1)^(2)`
`" "=(sin^(2)x+cos^(2)x)^(2)-2sin^(2)xcos^(2)x-(1)/(4)(a^(2)-1)^(2)`
`" "=1-2*(1)/(4)(a^(2)-1)^(2)-(1)/(4)(a^(2)-1)^(2)=(1)/(4)[4-3(a^(2)-1)^(2)]`
`(ii) abs(sinx-cosx)=sqrt((sinx-cosx)^(2))`
`" "=sqrt(sin^(2)x+cos^(2)x-2sinxcosx)`
`" "=sqrt(1-2(1)/(2)(a^(2)-1))=sqrt(1-a^(2)+1)=sqrt(2-a^(2))`