Question

if `sinx+cosx=a` find `sin^3(x)+cos^3(x)` in terms of `a`?

`Sinx+cosx=a`

`sin^3(x)+cos^3(x)=?`

Answer 1

`sin^3x+cos^3x = 1/2a(3-a^2)`

Explanation:

Consider `(sinx+cosx)^2`:

`(sinx+cosx)^2 =sin^2x+2sinxcosx+cos^2x`

Using `sinx+cosx=a` in the above we have;

`a^2 =sin^2x+cos^2x+2sinxcosx`
`\ \ \ =1+2sinxcosx`
`=> sinxcosx = (a^2-1)/2`

Consider now, `(sinx+cosx)^3`:

`(sinx+cosx)^3 =sin^3x+3sin^2xcosx+3sinxcos^2x+cos^3x`
`" " =sin^3x+cos^3x+3sinxcosx(sinx+cosx)`

Again, using `sinx+cosx=a` we have:

`a^3 =sin^3x+cos^3x+3asinxcosx`
`\ \ \ =sin^3x+cos^3x+3a * (a^2-1)/2`

`=> sin^3x+cos^3x = a^3 - 3a * (a^2-1)/2`
`" " = (2a^3 - 3a^3+3a)/2`
`" " = (3a-a^3)/2`
`" " = 1/2a(3-a^2)`

Answer 2

`sin^3x+cos^3x`

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

`=a(1-1/2xx2sinx cosx)`

`=a(1-1/2xx((sinx +cosx)^2-1)`

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

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

`=a/2(3-a^2)`