[tex]sin^5xcos^2x = (cos^2x-2cos^4x+cos^6x)sinx\ is\ proved[/tex]
Solution:
Given that we have to prove:
[tex]sin^5xcos^2x = (cos^2x-2cos^4x+cos^6x)sinx[/tex]
Let us first take the left hand side of equation
[tex]sin^5xcos^2x\\\\Split\ sin^5x\ as\ (sin^4x)(sinx)[/tex]
[tex](sin^4x)(sinx)(cos^2x)\\\\Which\ is\\\\(sinx)(sin^4x)(cos^2x) ------ eqn 1[/tex]
Now take the right side of equation
[tex](cos^2x-2cos^4x+cos^6x)sinx\\\\Take\ cos^2x\ as\ common\ term\ out\\\\(1-2cos^2x + cos^4x)cos^2x\ sinx[/tex]
[tex]By\ algebraic\ identity,\\\\(a-b)^2 = a^2-2ab + b^2[/tex]
[tex]Therefore,\\\\(1-cos^2x)^2cos^2x\ sinx\\\\We\ know\ that\\\\1-cos^2x = sin^2x\\\\Therefore\\\\sin^4xcos^2x sinx ----- eqn 2[/tex]
Eqn 1 = eqn 2
Therefore, L.H.S = R.H.S
[tex]sin^5xcos^2x = (cos^2x-2cos^4x+cos^6x)sinx[/tex]
Thus proved
