#(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#