Question 1022487
{{{sin^2(x)+sin^6(x)+sin^8(x) = rho}}}
==> {{{1+sin^2(x)+sin^6(x)(1+sin^2(x)) = 1 + rho}}}
<==> {{{(1+sin^2(x))(1+sin^6(x))= 1 + rho}}}

Now {{{cosx+cos^2(x) = 1}}}==> {{{cosx = 1-cos^2(x) = sin^2(x)}}}.

==> {{{(1+cosx)(1+cos^3(x))= 1 + rho}}}  after direct substitution.

<==> {{{1+cos^3(x) +cosx + cos^4(x) = 1+rho}}}
==> {{{cosx +cos^3(x) + cos^4(x) = rho}}}

Now {{{cosx+cos^2(x) = 1}}}==> {{{cos^3(x)+cos^4(x) = cos^2(x)}}} after multiplying both sides by {{{cos^2(x)}}}

==> {{{cosx+cos^2(x) = rho}}} after direct substitution.

==> {{{rho = 1}}} because of the given.

Therefore, {{{sin^2(x)+sin^6(x)+sin^8(x) = highlight(1)}}}