`(sin x)^4 / x + (cos x )^4 /y = 1/(x+y)` Then prove that `(sin x)^12 /x^5 + (cos x)^12 /y^5 = 1/(x+y)^5` ?

Solving for `y`

`sin^4(x)/x + cos^4(x)/y = 1/(x + y)` we obtain after some simplifications

`y = x (cos(x)/sin(x))^2 = x cot^2(x)`

Now, substituting back in

`(sin x)^12 /x^5 + (cos x)^12 /y^5 = 1/(x+y)^5`

we can easily verify that the equality is observed.

Another way is making

`y^5sin^12x+y^5cos^12x=((xy)/(x+y))^5` and using the previous result

`y sin^2x=x cos^2x->y^5 sin^10x=x^5cos^10x`

then

`y^5sin^12x+y^5sin^10xcos^2x=((xy)/(x+y))^5` so

`y^5sin^10x=((xy)/(x+y))^5` so

`sin^10x=(x/(x+xcos^2x/sin^2x))^5` et voila!

GIVEN
`sin^4x/x+cos^4x/y=1/(x+y)`

`=>((x+y)/x)sin^4x+((x+y)/y)cos^4x=1`

`=>(1+y/x)sin^4x+(1+x/y)cos^4x=1`

`=>(sin^2x+cos^2x)^2-2sin^2xcos^2x+(y/x)sin^4x+(x/y)cos^4x=1`

`=>1-2sin^2xcos^2x+(y/x)sin^4x+(x/y)cos^4x=1`

`=>-2sin^2xcos^2x+(y/x)sin^4x+(x/y)cos^4x=0`

`=>(sqrt(y/x)sin^2x-sqrt(x/y)cos^2x)^2=0`

`=>sqrt(y/x)sin^2x=sqrt(x/y)cos^2x`

`=>sin^2x/x=cos^2x/y`

By addendo

`=>sin^2x/x=cos^2x/y=(sin^2x+cos^2x)/(x+y)=1/(x+y)`

`=>sin^2x=x/(x+y)`

And

`=>cos^2x=y/(x+y)`

Now

`sin^12x/x^5+cos^12x/y^5`

`=(sin^2x)^6/x^5+(cos^2x)^6/y^5`

`=(x/(x+y))^6/x^5+(y/(x+y))^6/y^5`

`=x/(x+y)^6+y/(x+y)^6`

`=1/(x+y)^5`